For Use Only in 2013 – 2014 Pilot Program 96 Chapter 2 r Limits b. Create a graph that gives a more complete representation of f. 46–47. Steep secant lines a. Given the graph of f in the following figures, find the slope of the secant line that passes through 10, 02 and 1h, f 1h22 in terms of h, for h 7 0 and h 6 0. b. Evaluate the limit of the slope of the secant line found in part (a) as h S 0+ and h S 0-. What does this tell you about the line tangent to the curve at 10, 02? 46. f 1x2 = x y 20 15 1>3 10 y y⫽ 2000 50 ⫹ 100 x2 2 4 5 (h, h1/3) (0, 0) ⫺4 x h T 47. f 1x2 = x 2>3 x2/3 (h, h2/3) 49. f 1x2 = x 2 - 3x + 2 x 10 - x 9 50. g1x2 = 2 - ln x 2 51. h1x2 = ex 1x + 123 px 52. p1x2 = sec a b , for x 6 2 2 53. g1u2 = tan a (0, 0) x h 55. f 1x2 = T 48. Care with graphing The figure shows the graph of the function 2000 graphed in the window 3 -4, 44 * 30, 204. f 1x2 = 50 + 100x 2 xS0 pu b 10 1 1x sec x 54. q1s2 = p s - sin s 56. g1x2 = e1>x 57. Can a graph intersect a vertical asymptote? A common misconception is that the graph of a function never intersects its vertical asymptotes. Let 4 f 1x2 = W x - 1 x2 a. Evaluate lim+ f 1x2, lim- f 1x2, and lim f 1x2. xS0 if x 6 1 if x Ú 1 . Explain why x = 1 is a vertical asymptote of the graph of f and show that the graph of f intersects the line x = 1. QUICK CHECK ANSWERS 1. Answers will vary, but all graphs should have a vertical asymptote at x = 2. 2. - ∞; ∞ 3. As x S -4 + , x 6 0 and 1x + 42 7 0, so x1x + 42 S 0 through negative values. 1x - 121x - 22 4. lim = lim 1x - 12 = 1, which is not xS2 x - 2 xS2 an infinite limit, so x = 2 is not a vertical asymptote. ➤ xS0 x 49–56. Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. y T 0 Technology Exercises y ⫽ x1/3 y⫽ ⫺2 2.5 Limits at Infinity Limits at infinity—as opposed to infinite limits—occur when the independent variable becomes large in magnitude. For this reason, limits at infinity determine what is called the end behavior of a function. An application of these limits is to determine whether a system (such as an ecosystem or a large oscillating structure) reaches a steady state as time increases. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.5 Limits at Infinity Limits at Infinity and Horizontal Asymptotes lim f (x) q y x Consider the function f 1x2 = tan-1 x, whose domain is 1- ∞, ∞2 (Figure 2.30). As x becomes arbitrarily large (denoted x S ∞ ), f 1x2 approaches p>2, and as x becomes arbitrarily large in magnitude and negative (denoted x S - ∞ ), f 1x2 approaches -p>2. These limits are expressed as ᠬ q Horizontal asymptote f (x) tan1 x lim tan-1 x = x 2 97 xS∞ Horizontal asymptote p 2 and p lim tan-1 x = - . 2 x S -∞ The graph of f approaches the horizontal line y = p>2 as x S ∞ , and it approaches the horizontal line y = -p>2 as x S - ∞. These lines are called horizontal asymptotes. q lim f (x) q x ᠬ FIGURE 2.30 DEFINITION Limits at Infinity and Horizontal Asymptotes y If f 1x2 becomes arbitrarily close to a finite number L for all sufficiently large and positive x, then we write lim f (x) L x ᠬ L x lim f 1x2 = L. xS∞ y f (x) f (x) x f (x) x We say the limit of f 1x2 as x approaches infinity is L. In this case, the line y = L is a horizontal asymptote of f (Figure 2.31). The limit at negative infinity, lim f 1x2 = M, is defined analogously. When the limit exists, the horizontal xS - ∞ asymptote is y = M. M lim f (x) M x QUICK CHECK 1 Evaluate x>1x + 12 for x = 10, 100, and 1000. What is lim xS∞ x ? x +1 ➤ ᠬ FIGURE 2.31 EXAMPLE 1 Limits at infinity Evaluate the following limits. a. lim a2 + xS - ∞ ➤ The limit laws of Theorem 2.3 and the Squeeze Theorem apply if x S a is replaced with x S ∞ or x S - ∞. 10 b x2 b. lim a3 + xS∞ 3 sin x b 1x SOLUTION a. As x becomes large and negative, x 2 becomes large and positive; in turn, 10>x 2 approaches 0. By the limit laws of Theorem 2.3, y 10 x2 lim a2 + x S -∞ 10 10 b = lim 2 + lim a 2 b = 2 + 0 = 2. x S -∞ x S -∞ x x2 (++)++* c f (x) 2 equals 2 lim f (x) 2 lim f (x) 2 x x ᠬ ᠬ 2 FIGURE 2.32 10 b is also equal to 2. Therefore, the graph of y = 2 + 10>x 2 x2 approaches the horizontal asymptote y = 2 as x S ∞ and as x S - ∞ (Figure 2.32). Notice that lim a2 + xS∞ y2 2 equals 0 b. The numerator of sin x> 1x is bounded between -1 and 1; therefore, for x 7 0, x - 1 sin x 1 … … . 1x 1x 1x As x S ∞, 1x becomes arbitrarily large, which means that lim xS∞ -1 1 = lim = 0. x S ∞ 1x 1x It follows by the Squeeze Theorem (Theorem 2.5) that lim xS∞ Copyright © 2014 Pearson Education, Inc. sin x = 0. 1x For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 98 Limits Using the limit laws of Theorem 2.3, y 6 lim a3 + 3 sin x f (x) 3 x 3 sin x sin x b = lim 3 + 3 lim a b = 3. xS∞ xS∞ 1x (++)+1x +* e 5 xS∞ equals 3 4 equals 0 3 sin x approaches the horizontal asymptote y = 3 as x 1x becomes large (Figure 2.33). Note that the curve intersects its asymptote infinitely many times. Related Exercises 9–14 The graph of y = 3 + 2 lim f (x) 3 x 1 0 0 10 20 30 x 40 FIGURE 2.33 Infinite Limits at Infinity It is possible for a limit to be both an infinite limit and a limit at infinity. This type of limit occurs if f 1x2 becomes arbitrarily large in magnitude as x becomes arbitrarily large in magnitude. Such a limit is called an infinite limit at infinity and is illustrated by the function f 1x2 = x 3 (Figure 2.34). lim f (x) y ➤ 3 x ᠬ DEFINITION Infinite Limits at Infinity 1 If f 1x2 becomes arbitrarily large as x becomes arbitrarily large, then we write f (x) x3 lim f 1x2 = ∞. 1 1 1 lim f (x) x ᠬ FIGURE 2.34 xS∞ x The limits lim f 1x2 = - ∞, lim f 1x2 = ∞, and lim f 1x2 = - ∞ are xS∞ xS - ∞ defined similarly. xS - ∞ Infinite limits at infinity tell us about the behavior of polynomials for large-magnitude values of x. First, consider power functions f 1x2 = x n, where n is a positive integer. Figure 2.35 shows that when n is even, lim x n = ∞, and when n is odd, lim x n = ∞ xS { ∞ xS∞ and lim x n = - ∞. xS - ∞ x n 0 even: lim x n n 0 odd: y lim x n lim x n y 60 x y x6 y x4 x y x5 yx 7 20 y x3 40 3 2 1 20 y x2 3 2 1 1 2 3 x FIGURE 2.35 Copyright © 2014 Pearson Education, Inc. 20 2 3 x For Use Only in 2013 – 2014 Pilot Program 2.5 Limits at Infinity 99 It follows that reciprocals of power functions f 1x2 = 1>x n = x -n, where n is a positive integer, behave as follows: lim xS∞ 1 = lim x -n = 0. xn xS - ∞ From here, it is a short step to finding the behavior of any polynomial as x S { ∞. Let p1x2 = anx n + an - 1x n - 1 + g + a2x 2 + a1x + a0. We now write p in the equivalent form a0 an - 2 an - 1 + 2 + g + n ¢. x x x " S0 e p1x2 = x n °an + e S0 S0 Notice that as x becomes large in magnitude, all the terms in p except the first term approach zero. Therefore, as x S {∞, we see that p1x2 ≈ anx n. This means that as x S { ∞, the behavior of p is determined by the term anx n with the highest power of x. THEOREM 2.6 Limits at Infinity of Powers and Polynomials Let n be a positive integer and let p be the polynomial p1x2 = anx n + an - 1x n - 1 + g + a2x 2 + a1x + a0, where an ≠ 0. 1. lim x n = ∞ when n is even. xS { ∞ 2. lim x n = ∞ and lim x n = - ∞ when n is odd. xS∞ xS - ∞ 1 3. lim n = lim x -n = 0. S x {∞ x xS { ∞ 4. lim p1x2 = lim a n x n = ∞ or - ∞, depending on the degree of the xS { ∞ xS { ∞ polynomial and the sign of the leading coefficient an. EXAMPLE 2 Limits at infinity Evaluate the limits as x S {∞ of the following functions. a. p1x2 = 3x 4 - 6x 2 + x - 10 b. q1x2 = -2x 3 + 3x 2 - 12 SOLUTION a. We use the fact that the limit is determined by the behavior of the leading term: lim 13x 4 - 6x 2 + x - 102 = lim 3 x 4 = ∞. xS∞ b xS∞ S∞ Similarly, lim 13x 4 - 6x 2 + x - 102 = lim 3 x 4 = ∞. xS - ∞ S∞ b xS - ∞ b. Noting that the leading coefficient is negative, we have lim 1-2x 3 + 3x 2 - 122 = lim 1-2 x 32 = - ∞ xS∞ S∞ b xS∞ lim 1-2x 3 + 3x 2 - 122 = lim 1-2 x 32 = ∞. xS - ∞ xS - ∞ S -∞ b Describe the behavior of p1x2 = -3x 3 as x S ∞ and as x S - ∞. lim xS - ∞ Related Exercises 15–24 Copyright © 2014 Pearson Education, Inc. ➤ QUICK CHECK 2 1 = lim x -n = 0 and xn xS∞ ➤ For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 100 Limits End Behavior The behavior of polynomials as x S {∞ is an example of what is often called end behavior. Having treated polynomials, we now turn to the end behavior of rational, algebraic, and transcendental functions. EXAMPLE 3 End behavior of rational functions Determine the end behavior for the following rational functions. a. f 1x2 = 3x + 2 x2 - 1 b. g1x2 = 40x 4 + 4x 2 - 1 10x 4 + 8x 2 + 1 c. h1x2 = x 3 - 2x + 1 2x + 4 SOLUTION a. An effective approach for evaluating limits of rational functions at infinity is to divide both the numerator and denominator by x n, where n is the largest power appearing in the denominator. This strategy forces the terms corresponding to lower powers of x to approach 0 in the limit. In this case, we divide by x 2: approaches 0 b 3x + 2 3 2 + 2 x 3x + 2 0 x2 x = lim 2 lim 2 = lim = = 0. xS∞ x - 1 xS∞ x - 1 xS∞ 1 1 1 - 2 2 x x approaches 0 ➤ Recall that the degree of a polynomial is xS - ∞ the highest power of x that appears. y f (x) 3x 2 x2 1 b. Again we divide both the numerator and denominator by the largest power appearing in the denominator, which is x 4: 1 1 3x + 2 = 0, and thus the graph of f has the horizontal x2 - 1 asymptote y = 0. You should confirm that the zeros of the denominator are -1 and 1, which correspond to vertical asymptotes (Figure 2.36). In this example, the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator. A similar calculation gives lim x 1 lim f (x) 0 x ᠬ lim f (x) 0 x ᠬ 40x 4 4x 2 1 + - 4 4 4 40x + 4x - 1 x x x = lim lim x S ∞ 10x 4 + 8x 2 + 1 x S ∞ 10x 4 8x 2 1 + 4 + 4 4 x x x 4 2 Divide the numerator and denominator by x 4. approaches 0 approaches 0 y lim g(x) 4 xS∞ x 4 ᠬ 10 + b x ᠬ = lim lim g(x) 4 4 x2 8 x2 + 1 x4 1 x4 Simplify. b 40 + b b FIGURE 2.36 approaches 0 approaches 0 2 4 2 FIGURE 2.37 40x4 4x2 1 g(x) 10x4 8x2 1 2 4 x = 40 + 0 + 0 = 4. 10 + 0 + 0 Evaluate limits. Using the same steps (dividing each term by x 4), it can be shown that 40x 4 + 4x 2 - 1 = 4. This function has the horizontal asymptote y = 4 lim x S - ∞ 10x 4 + 8x 2 + 1 (Figure 2.37). Notice that the degree of the polynomial in the numerator equals the degree of the polynomial in the denominator. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.5 Limits at Infinity 101 c. We divide the numerator and denominator by the largest power of x appearing in the denominator, which is x, and then take the limit: x3 2x 1 + x x x x - 2x + 1 lim = lim xS∞ 2x + 4 xS∞ 2x 4 + x x 3 Divide numerator and denominator by x. b b - x2 2 + + 4 x b 2 constant = ∞. 1 x Simplify. b = lim xS∞ approaches 0 b arbitrarily large constant approaches 0 Take limits. As x S ∞, all the terms in this function either approach zero or are constant—except the x [email protected] in the numerator, which becomes arbitrarily large. Therefore, the limit of the x 3 - 2x + 1 function does not exist. Using a similar analysis, we find that lim = ∞. x S -∞ 2x + 4 These limits are not finite, and so the graph of the function has no horizontal asymptote. In this case, the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator. ➤ Related Exercises 25–34 The conclusions reached in Example 3 can be generalized for all rational functions. These results are summarized in Theorem 2.7 (Exercise 74). THEOREM 2.7 End Behavior and Asymptotes of Rational Functions p1x2 Suppose f 1x2 = is a rational function, where q1x2 p1x2 = amx m + am - 1x m - 1 + g + a2x 2 + a1x + a0 and q1x2 = bnx n + bn - 1x n - 1 + g + b2x 2 + b1x + b0, with am ≠ 0 and bn ≠ 0. QUICK CHECK 3 Use Theorem 2.7 to find the vertical and horizontal 10x asymptotes of y = . 3x - 1 a. Degree of numerator less than degree of denominator If m 6 n, then lim f 1x2 = 0, and y = 0 is a horizontal asymptote of f . xS { ∞ b. Degree of numerator equals degree of denominator If m = n, then lim f 1x2 = am >bn, and y = am >bn is a horizontal asymptote of f . xS { ∞ c. Degree of numerator greater than degree of denominator If m 7 n, then lim f 1x2 = ∞ or - ∞, and f has no horizontal asymptote. xS { ∞ d. Assuming that f 1x2 is in reduced form (p and q share no common factors), vertical asymptotes occur at the zeros of q. Copyright © 2014 Pearson Education, Inc. ➤ For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 102 Limits Although it isn’t stated explicitly, Theorem 2.7 implies that a rational function can have at most one horizontal asymptote, and whenever there is a horizontal asymptote, p1x2 p1x2 lim = lim . The same cannot be said of other functions, as the next examples x S ∞ q1x2 x S - ∞ q1x2 show. EXAMPLE 4 End behavior of an algebraic function Examine the end behavior of 10x 3 - 3x 2 + 8 f 1x2 = 225x 6 + x 4 + 2 . SOLUTION The square root in the denominator forces us to revise the strategy used with ra- tional functions. First, consider the limit as x S ∞. The highest power of the polynomial in the denominator is 6. However, the polynomial is under a square root, so effectively, the highest power in the denominator is 2x 6 = x 3. Dividing the numerator and denominator by x 3, for x 7 0, the limit is evaluated as follows: 10x - 3x + 8 3 lim xS∞ 2 225x 6 + x 4 + 2 10x 3 3x 2 8 - 3 + 3 3 x x x = lim xS∞ Divide by 2x 6 = x 3. 25x 6 x4 2 + 6 + 6 6 B x x x approaches 0 approaches 0 = lim 1 x2 25 + + 8 x3 2 x6 b A + approaches 0 = ➤ Recall that approaches 0 10 = 2. 125 Evaluate limits. As x S - ∞, x 3 is negative, so we divide numerator and denominator by 2x 6 = -x 3 (which is positive): x if x Ú 0 - x 3 if x 6 0. Because x is negative as x S - ∞, we have 2x 6 = - x 3. 10x - 3x + 8 3 3 lim xS - ∞ 2 225x + x + 2 6 4 = lim xS - ∞ 10x 3 3x 2 8 + 3 -x -x 3 -x 3 6 25x x 2 + 6 + 6 6 B x x x approaches 0 approaches 0 = lim xS - ∞ 6 10x3 3x2 8 f (x) 兹25x6 x4 2 = - 2 lim f (x) 2 x ᠬ 1 2 lim f (x) 2 ᠬ FIGURE 2.38 - 1 x2 + approaches 0 1 x 3 x b A 25 + b b -10 + y x Divide by 2x 6 = -x 3 7 0. 4 10 = -2. 125 8 x3 2 x6 Simplify. b Therefore, approaches 0 Evaluate limits. The limits reveal two asymptotes, y = 2 and y = -2. Observe that the graph crosses both horizontal asymptotes (Figure 2.38). Related Exercises 35–38 ➤ x if x Ú 0 2x 2 = x = b - x if x 6 0. 2x 6 = x 3 = b Simplify. b xS∞ b b 3 x 10 - EXAMPLE 5 End behavior of transcendental functions Determine the end behavior of the following transcendental functions. a. f 1x2 = ex and g 1x2 = e-x b. h1x2 = ln x Copyright © 2014 Pearson Education, Inc. c. f 1x2 = cos x For Use Only in 2013 – 2014 Pilot Program 2.5 Limits at Infinity 103 SOLUTION y lim ex x ᠬ f (x) ex a. The graph of f 1x2 = ex (Figure 2.39) makes it clear that as x S ∞, ex increases without bound. All exponential functions bx with b 7 1 behave this way, because raising a number greater than 1 to ever-larger powers produces numbers that increase without bound. The figure also suggests that as x S - ∞, the graph of ex approaches the horizontal asymptote y = 0. This claim is confirmed analytically by recognizing that lim ex 0 1 x = 0. xS∞ e lim ex = lim e-x = lim x ᠬ xS - ∞ x 0 x xS∞ xS - ∞ lim e-x = 0 and lim e-x = ∞. FIGURE 2.39 xS∞ f (x) ex yx lim ln x x h(x) f 1(x) ln x 1 x 1 Reflection of y e x across line y x x lim ln x 0 xS - ∞ b. The domain of ln x is 5 x: x 7 06 , so we evaluate lim+ ln x and lim ln x to determine xS0 xS0 It is not obvious whether the graph of ln x approaches a horizontal asymptote or whether the function grows without bound as x S ∞. Furthermore, the numerical evidence (Table 2.9) is inconclusive because ln x increases very slowly. The inverse relation between ex and ln x is again useful. The fact that the domain of ex is 1 - ∞,∞2 implies that the range of ln x is also 1- ∞,∞2. Therefore, the values of ln x lie in the interval 1- ∞, ∞2, and it follows that lim ln x = ∞. xS∞ c. The cosine function oscillates between -1 and 1 as x approaches infinity (Figure 2.41). Therefore, lim cos x does not exist. For the same reason, lim cos x does not exist. xS∞ xS - ∞ Table 2.9 x 10 105 1010 1050 1099 T ∞ xS∞ end behavior. For the first limit, recall that ln x is the inverse of ex (Figure 2.40), and the graph of ln x is a reflection of the graph of e x across the line y = x. The horizontal asymptote 1y = 02 of ex is also reflected across y = x, becoming a vertical asymptote 1x = 02 for ln x. These observations imply that lim+ ln x = - ∞. y ln x 2.302 11.513 23.026 115.129 227.956 T ??? f (x) cos x 1 x 1 lim cos x does not exist. x FIGURE 2.41 lim cos x does not exist. x Related Exercises 39–44 ➤ y FIGURE 2.40 xS∞ Therefore, lim e = ∞ and lim e = 0. Because e-x = 1>ex, it follows that x The end behavior of exponential and logarithmic functions are important in upcoming work. We summarize these results in the following theorem. End Behavior of ex, e-x, and ln x The end behavior for ex and e-x on 1- ∞, ∞2 and ln x on 10, ∞2 is given by the following limits: THEOREM 2.8 QUICK CHECK 4 How do the functions e10x and e-10x behave as x S ∞ and as x S - ∞? lim e x = ∞ and lim e -x = 0 and lim ln x = - ∞ and xS∞ xS∞ xS0 + Copyright © 2014 Pearson Education, Inc. lim e x = 0, xS - ∞ lim e -x = ∞, xS - ∞ lim ln x = ∞. xS∞ ➤ For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 104 Limits SECTION 2.5 EXERCISES Review Questions 35–38. Algebraic functions Evaluate lim f 1x2 and lim f 1x2 for the xS∞ 1. Explain the meaning of lim f 1x2 = 10. 2. What is a horizontal asymptote? f 1x2 Determine lim if f 1x2 S 100,000 and g1x2 S ∞ as x S ∞. x S ∞ g1x2 35. f 1x2 = 4. Describe the end behavior of g1x2 = e-2x. 36. f 1x2 = 5. Describe the end behavior of f 1x2 = - 2x 3. 6. The text describes three cases that arise when examining the end behavior of a rational function f 1x2 = p1x2>q1x2. Describe the end behavior associated with each case. 3. xS - ∞ following functions. Then give the horizontal asymptote(s) of f (if any). xS - ∞ 7. Evaluate lim ex, lim ex, and lim e-x. 8. Use a sketch to find the end behavior of f 1x2 = ln x. 37. f 1x2 = 4x 3 + 1 2x + 216x 6 + 1 3 2x 2 + 1 2x + 1 3 2x 6 + 8 4x 2 + 23x 4 + 1 38. f 1x2 = 4x 1 3x - 29x 2 + 1 2 Basic Skills 39–44. Transcendental functions Determine the end behavior of the following transcendental functions by evaluating appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. 9–14. Limits at infinity Evaluate the following limits. 39. f 1x2 = - 3e-x 40. f 1x2 = 2x 41. f 1x2 = 1 - ln x 42. f 1x2 = ln x 43. f 1x2 = sin x 44. f 1x2 = 9. xS - ∞ xS∞ 10 lim a3 + 2 b xS∞ x 11. lim uS ∞ xS∞ 1 10 10. lim a5 + + 2 b x xS∞ x cos u u2 12. lim xS∞ cos x 5 x S ∞ 1x 14. 13. lim 3 + 2x + 4x 2 x2 100 sin4 x 3 b + x x2 lim a5 + xS - ∞ 15–24. Infinite limits at infinity Determine the following limits. 15. lim x 12 16. xS∞ 17. lim x -6 18. xS∞ 19. lim 13x xS∞ 21. 12 - 9x 2 7 lim 1- 3x 16 + 22 xS - ∞ 23. lim 1-12x -52 xS∞ 20. 22. 24. 45. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The graph of a function can never cross one of its horizontal asymptotes. b. A rational function f can have both lim f 1x2 = L and c. The graph of any function can have at most two horizontal asymptotes. lim x - 11 x S -∞ lim 13x 7 + x 22 xS - ∞ lim 2x -8 xS - ∞ lim 12x -8 + 4x 32 xS - ∞ 25–34. Rational functions Evaluate lim f 1x2 and lim f 1x2 for the xS - ∞ following rational functions. Then give the horizontal asymptote of f (if any). 3x 2 - 7 x 2 + 5x 46–55. Horizontal and vertical asymptotes a. Evaluate lim f 1x2 and lim f 1x2, and then identify any xS - ∞ xSa 2 xSa x - 4x + 3 46. f 1x2 = x - 1 48. f 1x2 = 216x 4 + 64x 2 + x 2 2x 2 - 4 49. f 1x2 = 3x 4 + 3x 3 - 36x 2 x 4 - 25x 2 + 144 26. f 1x2 = 27. f 1x2 = 6x 2 - 9x + 8 3x 2 + 2 28. f 1x2 = 4x 2 - 7 8x + 5x + 2 29. f 1x2 = 3x 3 - 7 x 4 + 5x 2 30. f 1x2 = x4 + 7 x5 + x2 - x 51. f 1x2 = 31. f 1x2 = 2x + 1 3x 4 - 2 32. f 1x2 = 12x 8 - 3 3x 8 - 2x 7 52. f 1x2 = 33. f 1x2 = 40x 5 + x 2 16x 4 - 2x 34. f 1x2 = -x 3 + 1 2x + 8 53. f 1x2 = 2 xS∞ horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote x = a, evaluate lim- f 1x2 and lim+ f 1x2. 4x 20x + 1 25. f 1x2 = xS∞ xS - ∞ x S -∞ xS∞ Further Explorations lim f 1x2 = ∞ . lim 3 x 11 50 e2x 47. f 1x2 = 50. f 1x2 = 16x 2 1 4x 2 - 216x 4 + 1 2 x2 - 9 x1x - 32 x - 1 x 2>3 - 1 2x 2 + 2x + 6 - 3 x - 1 Copyright © 2014 Pearson Education, Inc. 2x 3 + 10x 2 + 12x x 3 + 2x 2 For Use Only in 2013 – 2014 Pilot Program 2.5 Limits at Infinity 54. f 1x2 = 105 61. lim+ f 1x2 = ∞ , lim- f 1x2 = - ∞ , lim f 1x2 = 1, 1 - x2 xS0 x1x + 12 xS0 lim f 1x2 = -2 xS - ∞ 55. f 1x2 = 2 x - 2 x - 1 62. Asymptotes Find the vertical and horizontal asymptotes of f 1x2 = e1>x. 56–59. End behavior for transcendental functions 56. The central branch of f 1x2 = tan x is shown in the figure. a. Evaluate lim -tan x and x S p>2 lim tan x. Are these x S -p>2 + infinite limits or limits at infinity? b. Sketch a graph of g1x2 = tan-1x by reflecting the graph of f over the line y = x, and use it to evaluate lim tan-1x and xS∞ lim tan-1 x. xS - ∞ y xS∞ 63. Asymptotes Find the vertical and horizontal asymptotes of cos x + 21x . f 1x2 = 1x Applications 64–69. Steady states If a function f represents a system that varies in time, the existence of lim f 1t2 means that the system reaches a steady tS ∞ state (or equilibrium). For the following systems, determine if a steady state exists and give the steady-state value. f (x) tan x 64. The population of a bacteria culture is given by p1t2 = 2500 . t + 1 65. The population of a culture of tumor cells is given by p1t2 = 3500t . t + 1 1 q q 1 66. The amount of drug (in milligrams) in the blood after an IV tube is inserted is m1t2 = 20011 - 2-t2. x 67. The value of an investment in dollars is given by v1t2 = 1000e 0.065t. 68. The population of a colony of squirrels is given by 1500 p1t2 = . 3 + 2e-0.1t T 57. Graph y = sec -1 x and evaluate the following limits using the graph. Assume the domain is 5 x: x Ú 16. a. lim sec -1 x xS∞ b. lim sec -1 x xS - ∞ 58. The hyperbolic cosine function, denoted cosh x, is used to model the shape of a hanging cable (a telephone wire, for example). It is ex + e-x . defined as cosh x = 2 a. Determine its end behavior by evaluating lim cosh x and xS∞ lim cosh x. xS - ∞ b. Evaluate cosh 0. Use symmetry and part (a) to sketch a plausible graph for y = cosh x. 59. The hyperbolic sine function is defined as sinh x = ex - e-x . 2 a. Determine its end behavior by evaluating lim sinh x and xS∞ lim sinh x. xS - ∞ b. Evaluate sinh 0. Use symmetry and part (a) to sketch a plausible graph for y = sinh x. 60–61. Sketching graphs Sketch a possible graph of a function f that satisfies all the given conditions. Be sure to identify all vertical and horizontal asymptotes. 60. f 1- 12 = - 2, f 112 = 2, f 102 = 0, lim f 1x2 = 1, lim f 1x2 = - 1 69. The amplitude of an oscillator is given by a1t2 = 2 a t + sin t b. t 70–73. Looking ahead to sequences A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence 5 2, 4, 6, 8, c6 is specified by the function f 1n2 = 2n, where n = 1, 2, 3, c . The limit of such a sequence is lim f 1n2, nS ∞ provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences, or state that the limit does not exist. 4 4 2 4 70. e 4, 2, , 1, , , cf, which is defined by f 1n2 = , for n 3 5 3 n = 1, 2, 3, c 1 2 3 n - 1 71. e 0, , , , c f, which is defined by f 1n2 = , n 2 3 4 for n = 1, 2, 3, c n2 1 4 9 16 , 72. e , , , , cf, which is defined by f 1n2 = 2 3 4 5 n + 1 for n = 1, 2, 3, c n + 1 3 4 5 73. e 2, , , , c f, which is defined by f 1n2 = , 4 9 16 n2 for n = 1, 2, 3, c xS∞ xS - ∞ Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Limits Additional Exercises T p1x2 74. End behavior of a rational function Suppose f 1x2 = q1x2 is a rational function, where p1x2 = a mx m + a m - 1x m - 1 + g + a 2x 2 + a 1x + a 0, q1x2 = b nx n + b n - 1x n - 1 + g + b 2x 2 + b 1x + b 0, a m ≠ 0, and b n ≠ 0. am a. Prove that if m = n, then lim f 1x2 = . xS { ∞ bn QUICK CHECK ANSWERS 1. 10>11, 100>101, 1000>1001, 1 2. p1x2 S - ∞ as x S ∞ and p1x2 S ∞ as x S - ∞ 3. Horizontal 1 asymptote is y = 10 3 ; vertical asymptote is x = 3 . 4. lim e 10x = ∞, lim e 10x = 0, lim e -10x = 0, b. Prove that if m 6 n, then lim f 1x2 = 0. xS { ∞ 75–76. Limits of exponentials Evaluate lim f 1x2 and lim f 1x2. xS∞ xS∞ x S -∞ lim e -10x = ∞ Then state the horizontal asymptote(s) of f . Confirm your findings by plotting f . 75. f 1x2 = 2e x + 3e 2x e 2x + e 3x 76. f 1x2 = xS - ∞ xS∞ xS - ∞ 3e x + e - x ex + e - x 2.6 Continuity The graphs of many functions encountered in this text contain no holes, jumps, or breaks. For example, if L = f 1t2 represents the length of a fish t years after it is hatched, then the length of the fish changes gradually as t increases. Consequently, the graph of L = f 1t2 contains no breaks (Figure 2.42a). Some functions, however, do contain abrupt changes in their values. Consider a parking meter that accepts only quarters and each quarter buys 15 minutes of parking. Letting c1t2 be the cost (in dollars) of parking for t minutes, the graph of c has breaks at integer multiples of 15 minutes (Figure 2.42b). 25 1.25 Cost (dollars) y Length (in) L L ⫽ f (t) 5 1.00 y ⫽ c(t) 0.75 0.50 0.25 0 1 2 3 4 t 0 15 30 45 60 t Time (min) (b) Time (yr) (a) FIGURE 2.42 QUICK CHECK 1 For what values of t in 10, 602 does the graph of y = c1t2 in Figure 2.42b have a discontinuity? ➤ T 77. Subtle asymptotes Use analytical methods to identify all the ln 19 - x 22 . Then confirm your results by asymptotes of f 1x2 = 2e x - e - x locating the asymptotes using a graphing calculator. ➤ Chapter 2 r 106 Informally, we say that a function f is continuous at a if the graph of f contains no holes or breaks at a (that is, if the graph near a can be drawn without lifting the pencil). If a function is not continuous at a, then a is a point of discontinuity. Continuity at a Point This informal description of continuity is sufficient for determining the continuity of simple functions, but it is not precise enough to deal with more complicated functions such as h1x2 = W x sin 1 x 0 Copyright © 2014 Pearson Education, Inc. if x ≠ 0 if x = 0. For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity 107 It is difficult to determine whether the graph of h has a break at 0 because it oscillates rapidly as x approaches 0 (Figure 2.43). We need a better definition. y 0.2 Is h continuous at x 0? 0.1 0.2 x 0.2 0.1 h (x) 0.2 x sin 0 1 x if x 0 if x 0 FIGURE 2.43 DEFINITION Continuity at a Point A function f is continuous at a if lim f 1x2 = f 1a2. If f is not continuous at a, then a xSa is a point of discontinuity. There is more to this definition than first appears. If lim f 1x2 = f 1a2, then f 1a2 and xSa lim f 1x2 must both exist, and they must be equal. The following checklist is helpful in xSa determining whether a function is continuous at a. Continuity Checklist In order for f to be continuous at a, the following three conditions must hold. 1. f 1a2 is defined 1a is in the domain of f 2. 2. lim f 1x2 exists. xSa y 3. lim f 1x2 = f 1a2 1the value of f equals the limit of f at a2. xSa 5 y f (x) If any item in the continuity checklist fails to hold, the function fails to be continuous at a. From this definition, we see that continuity has an important practical consequence: 3 If f is continuous at a, then lim f 1x2 = f 1a2, and direct substitution may be used to 1 0 1 3 5 7 FIGURE 2.44 ➤ In Example 1, the discontinuities at x = 1 and x = 2 are called removable discontinuities because they can be removed by redefining the function at these points (in this case f 112 = 3 and f 122 = 1). The discontinuity at x = 3 is called a jump discontinuity. The discontinuity at x = 5 is called an infinite discontinuity. These terms are discussed in Exercises 95–101. x evaluate lim f 1x2. xSa xSa EXAMPLE 1 Points of discontinuity Use the graph of f in Figure 2.44 to identify values of x on the interval 10, 72 at which f has a discontinuity. SOLUTION The function f has discontinuities at x = 1, 2, 3, and 5 because the graph contains holes or breaks at each of these locations. These claims are verified using the continuity checklist. r f 112 is not defined. r f 122 = 3 and lim f 1x2 = 1. Therefore, f 122 and lim f 1x2 exist but are not equal. xS2 Copyright © 2014 Pearson Education, Inc. xS2 For Use Only in 2013 – 2014 Pilot Program Limits r lim f 1x2 does not exist because the left-sided limit lim- f 1x2 = 2 differs from the xS3 xS3 right-sided limit lim+ f 1x2 = 1. xS3 r Neither lim f 1x2 nor f 152 exists. Related Exercises 9–12 xS5 EXAMPLE 2 Identifying discontinuities Determine whether the following functions are continuous at a. Justify each answer using the continuity checklist. 3x 2 + 2x + 1 ; x - 1 3x 2 + 2x + 1 b. g1x2 = ; x - 1 1 if x x sin x c. h1x2 = W 0 if x a. f 1x2 = a = 1 a = 2 ≠ 0 ;a = 0 = 0 SOLUTION a. The function f is not continuous at 1 because f 112 is undefined. b. Because g is a rational function and the denominator is nonzero at 2, it follows by Theorem 2.3 that lim g1x2 = g122 = 17. Therefore, g is continuous at 2. xS2 c. By definition, h102 = 0. In Exercise 55 of Section 2.3, we used the Squeeze Theorem 1 to show that lim x sin = 0. Therefore, lim h1x2 = h102, which implies that h is S x x 0 xS0 continuous at 0. Related Exercises 13–20 ➤ Chapter 2 r ➤ 108 The following theorems make it easier to test various combinations of functions for continuity at a point. THEOREM 2.9 Continuity Rules If f and g are continuous at a, then the following functions are also continuous at a. Assume c is a constant and n 7 0 is an integer. a. f + g c. cf e. f >g, provided g1a2 ≠ 0 b. f - g d. fg f. 1 f 1x22n To prove the first result, note that if f and g are continuous at a, then lim f 1x2 = f 1a2 xSa and lim g1x2 = g1a2. From the limit laws of Theorem 2.3, it follows that xSa lim 1 f 1x2 + g1x22 = f 1a2 + g1a2. xSa Therefore, f + g is continuous at a. Similar arguments lead to the continuity of differences, products, quotients, and powers of continuous functions. The next theorem is a direct consequence of Theorem 2.9. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity THEOREM 2.10 109 Polynomial and Rational Functions a. A polynomial function is continuous for all x. p b. A rational function (a function of the form , where p and q are polynomials) q is continuous for all x for which q1x2 ≠ 0. EXAMPLE 3 Applying the continuity theorems For what values of x is the function 20 f 1x2 = 2 6 x x continuous? x 2 - 7x + 12 SOLUTION a. Because f is rational, Theorem 2.10b implies it is continuous for all x at which the denominator is nonzero. The denominator factors as 1x - 321x - 42, so it is zero at x = 3 and x = 4. Therefore, f is continuous for all x except x = 3 and x = 4 (Figure 2.45). Related Exercises 21–26 20 Continuous everywhere except x 3 and x 4 FIGURE 2.45 ➤ y x f (x) 2 x 7x 12 The following theorem allows us to determine when a composition of two functions is continuous at a point. Its proof is informative and is outlined in Exercise 102. THEOREM 2.11 Continuity of Composite Functions at a Point If g is continuous at a and f is continuous at g1a2, then the composite function f ∘ g is continuous at a. Evaluate lim 2x 2 + 9 and 2lim 1x 2 + 92. xS4 xS4 How do these results illustrate that the order of a function evaluation and a limit may be switched for continuous functions? Theorem 2.11 is useful because it allows us to conclude that the composition of two continuous functions is continuous at a point. For example, the composite function 3 x a b is continuous for all x ≠ 1. The theorem also says that under the stated condix - 1 tions on f and g, the limit of their composition is evaluated by direct substitution; that is, lim f 1g1x22 = f 1g1a22. ➤ xSa EXAMPLE 4 Limit of a composition Evaluate lim a xS0 x 4 - 2x + 2 10 b . x 6 + 2x 4 + 1 x 4 - 2x + 2 is continuous for all x because its x 6 + 2x 4 + 1 x 4 - 2x + 2 10 denominator is always positive (Theorem 2.10b). Therefore, a 6 b , which x + 2x 4 + 1 is the composition of the continuous function f 1x2 = x 10 and a continuous rational function, is continuous for all x by Theorem 2.11. By direct substitution, SOLUTION The rational function lim a xS0 x 4 - 2x + 2 10 04 - 2 # 0 + 2 10 b = a 6 b = 210 = 1024. 6 4 x + 2x + 1 0 + 2 # 04 + 1 Related Exercises 27–30 ➤ QUICK CHECK 2 Closely related to Theorem 2.11 are two results dealing with limits of composite functions; they are used frequently in upcoming chapters. We present these two results—one a more general version of the other—in a single theorem. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Limits THEOREM 2.12 Limits of Composite Functions 1. If g is continuous at a and f is continuous at g1a2, then 1 2 lim f 1g1x22 = f lim g1x2 . xSa xSa 2. If lim g1x2 = L and f is continuous at L, then xSa 1 2 lim f 1g1x22 = f lim g1x2 . xSa xSa Proof: The first statement follows directly from Theorem 2.11, which states that lim f 1g1x22 = f 1g1a22. If g is continuous at a, then lim g1x2 = g1a2, and it follows that xSa xSa 1 2 lim f 1g1x22 = f 1g1a22 = f lim g1x2 . xSa b xSa lim g1x2 xSa The proof of the second statement relies on the formal definition of a limit, which is discussed in Section 2.7. Both statements of Theorem 2.12 justify interchanging the order of a limit and a function evaluation. By the second statement, the inner function of the composition needn’t be continuous at the point of interest, but it must have a limit at that point. EXAMPLE 5 Limits of composite functions Evaluate the following limits. a. lim 22x 2 - 1 b. lim cos a x S -1 xS2 x2 - 4 b x - 2 SOLUTION a. We show later in this section that 1x is continuous for x Ú 0. The inner function of the composite function 22x 2 - 1 is 2x 2 - 1 and it is continuous and positive at -1. By the first statement of Theorem 2.12, lim 22x 2 - 1 = lim 2x 2 - 1 = 11 = 1. A x S -1 e x S -1 1 b. We show later in this section that cos x is continuous at all points of its domain. The x2 - 4 x2 - 4 inner function of the composite function cos a b is , which is not x - 2 x - 2 continuous at 2. However, lim a xS2 1x - 221x + 22 x2 - 4 b = lim = lim 1x + 22 = 4. x - 2 xS2 x - 2 xS2 Therefore, by the second statement of Theorem 2.12, lim cos a g xS2 x2 - 4 x2 - 4 b = cos a lim a b b = cos 4 ≈ -0.654. x - 2 xS2 x - 2 4 Related Exercises 31–34 Copyright © 2014 Pearson Education, Inc. ➤ Chapter 2 r ➤ 110 For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity y 111 Continuity on an Interval Continuous on [a, b) A function is continuous on an interval if it is continuous at every point in that interval. Consider the functions f and g whose graphs are shown in Figure 2.46. Both these functions are continuous for all x in 1a, b2, but what about the endpoints? To answer this question, we introduce the ideas of left-continuity and right-continuity. y f (x) DEFINITION Continuity at Endpoints O a A function f is continuous from the left (or left-continuous) at a if lim- f 1x2 = f 1a2 x b xSa (a) and f is continuous from the right (or right-continuous) at a if lim+ f 1x2 = f 1a2. xSa y Continuous on (a, b] Combining the definitions of left-continuous and right-continuous with the definition of continuity at a point, we define what it means for a function to be continuous on an interval. y g(x) DEFINITION Continuity on an Interval O a A function f is continuous on an interval I if it is continuous at all points of I. If I contains its endpoints, continuity on I means continuous from the right or left at the endpoints. x b (b) FIGURE 2.46 To illustrate these definitions, consider again the functions in Figure 2.46. In Figure 2.46a, f is continuous from the right at a because lim+ f 1x2 = f 1a2; but it is not continuous from xSa the left at b because f 1b2 is not defined. Therefore, f is continuous on the interval 3a, b2. The behavior of the function g in Figure 2.46b is the opposite: It is continuous from the left at b, but it is not continuous from the right at a. Therefore, g is continuous on 1a, b4 . QUICK CHECK 3 Modify the graphs of the functions f and g in Figure 2.46 to obtain functions that are continuous on 3a, b4 . ➤ EXAMPLE 6 Intervals of continuity Determine the intervals of continuity for f 1x2 = e y SOLUTION This piecewise function consists of two polynomials that describe a parabola 10 y f (x) x 2 + 1 if x … 0 3x + 5 if x 7 0. and a line (Figure 2.47). By Theorem 2.10, f is continuous for all x ≠ 0. From its graph, it appears that f is left-continuous at 0. This observation is verified by noting that Continuous on (0, ) lim f 1x2 = lim- 1x 2 + 12 = 1, x S 0- 2 FIGURE 2.47 2 which means that lim- f 1x2 = f 102. However, because Left-continuous at x 0 2 xS0 x lim f 1x2 = lim+ 13x + 52 = 5 ≠ f 102, x S 0+ xS0 we see that f is not right-continuous at 0. Therefore, we can also say that f is continuous on 1 - ∞, 04 and on 10, ∞2. Related Exercises 35–40 Copyright © 2014 Pearson Education, Inc. ➤ Continuous on (, 0] xS0 For Use Only in 2013 – 2014 Pilot Program 112 Chapter 2 r Limits Functions Involving Roots Recall that Limit Law 7 of Theorem 2.3 states 3 4 lim 3 f 1x24 n>m = lim f 1x2 n>m, xSa xSa provided f 1x2 Ú 0, for x near a, if m is even and n>m is reduced. Therefore, if m is odd and f is continuous at a, then 3 f 1x24 n>m is continuous at a, because 3 4 lim 3 f 1x24 n>m = lim f 1x2 xSa xSa n>m = 3 f 1a24 n>m. When m is even, the continuity of 3f 1x24 n>m must be handled more carefully because this function is defined only when f 1x2 Ú 0. Exercise 59 of Section 2.7 establishes an important fact: If f is continuous at a and f 1a2 7 0, then f is positive for all values of x in the domain sufficiently close to a. Combining this fact with Theorem 2.11 (the continuity of composite functions), it follows that 3 f 1x24 n>m is continuous at a provided f 1a2 7 0. At points where f 1a2 = 0, the behavior of 3 f 1x24 n>m varies. Often we find that 3 f 1x24 n>m is left- or right-continuous at that point, or it may be continuous from both sides. THEOREM 2.13 Continuity of Functions with Roots Assume that m and n are positive integers with no common factors. If m is an odd integer, then 3 f 1x24 n>m is continuous at all points at which f is continuous. If m is even, then 3 f 1x24 n>m is continuous at all points a at which f is continuous and f 1a2 7 0. EXAMPLE 6 Continuity with roots For what values of x are the following functions Continuous on [3, 3] continuous? y a. g1x2 = 29 - x 2 4 SOLUTION g(x) 兹9 x2 2 3 Right-continuous at x 3 x Left-continuous at x 3 xS3 that g is left-continuous at 3. Similarly, g is right-continuous at -3 because lim + 29 - x 2 = 0 = g1-32. Therefore, g is continuous on 3 -3, 34 . FIGURE 2.48 xS - 3 b. The polynomial x 2 - 2x + 4 is continuous for all x by Theorem 2.10a. Because f involves an odd root (m = 3, n = 2 in Theorem 2.13), f is continuous for all x. QUICK CHECK 4 On what interval is continuous? On what f 1x2 = x interval is f 1x2 = x 2>5 continuous? a. The graph of g is the upper half of the circle x 2 + y 2 = 9 (which can be verified by solving x 2 + y 2 = 9 for y). From Figure 2.48, it appears that g is continuous on 3 -3, 34 . To verify this fact, note that g involves an even root 1m = 2, n = 1 in Theorem 2.13). If -3 6 x 6 3, then 9 - x 2 7 0 and by Theorem 2.13, g is continuous for all x on 1-3, 32. At the right endpoint, lim- 29 - x 2 = 0 = g132 by Limit Law 7, which implies 1>4 Related Exercises 41–50 ➤ 3 b. f 1x2 = 1x 2 - 2x + 422>3 ➤ Continuity of Transcendental Functions The understanding of continuity that we have developed with algebraic functions may now be applied to transcendental functions. Trigonometric Functions In Example 8 of Section 2.3, we used the Squeeze Theorem to show that lim sin x = 0 and lim cos x = 1. Because sin 0 = 0 and cos 0 = 1, these xS0 xS0 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity 113 limits imply that sin x and cos x are continuous at 0. The graph of y = sin x (Figure 2.49) suggests that lim sin x = sin a for any value of a, which means that sin x is continuous xSa everywhere. The graph of y = cos x also indicates that cos x is continuous for all x. Exercise 105 outlines a proof of these results. With these facts in hand, we appeal to Theorem 2.9e to discover that the remaining trigonometric functions are continuous on their domains. For example, because sec x = 1>cos x, the secant function is continuous for all x for which cos x ≠ 0 (for all x except odd multiples of p>2) (Figure 2.50). Likewise, the tangent, cotangent, and cosecant functions are continuous at all points of their domains. y ⫽ sec x y y 3 y ⫽ sin x 1 sin a (a, sin a) ... sin x ⫺w sin a ⫺q q w x ⫺3 As x FIGURE 2.49 f (x) 4x 43 = 4 # 4 # 4 = 64; 4-2 = (0, 1) x FIGURE 2.51 sec x is continuous at all points of its domain. FIGURE 2.50 Exponential Functions The continuity of exponential functions of the form f 1x2 = b x, with 0 6 b 6 1 or b 7 1, raises an important question. Consider the function f 1x2 = 4x (Figure 2.51). Evaluating f is routine if x is rational: y Exponential functions are defined for all real numbers and are continuous on (, ), as shown in Chapter 6 x a a... 1 1 1 = ; 43>2 = 243 = 8; and 4-1>3 = 3 . 16 42 24 But what is meant by 4x when x is an irrational number, such as 12? In order for f 1x2 = 4x to be continuous for all real numbers, it must also be defined when x is an irrational number. Providing a working definition for an expression such as 412 requires mathematical results that don’t appear until Chapter 6. Until then, we assume without proof that the domain of f 1x2 = b x is the set of all real numbers and that f is continuous at all points of its domain. Inverse Functions Suppose a function f is continuous and one-to-one on an interval I. Reflecting the graph of f through the line y = x generates the graph of f -1. The reflection process introduces no discontinuities in the graph of f -1, so it is plausible (and indeed, true) that f -1 is continuous on the interval corresponding to I. We state this fact without a formal proof. THEOREM 2.14 Continuity of Inverse Functions If a continuous function f has an inverse on an interval I, then its inverse f -1 is also continuous (on the interval consisting of the points f 1x2, where x is in I). Because all the trigonometric functions are continuous on their domains, they are also continuous when their domains are restricted for the purpose of defining inverse functions. Therefore, by Theorem 2.14, the inverse trigonometric functions are continuous at all points of their domains. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 114 Limits Logarithmic functions of the form f 1x2 = logb x are continuous at all points of their domains for the same reason: They are inverses of exponential functions, which are one-to-one and continuous. Collecting all these facts together, we have the following theorem. THEOREM 2.15 Continuity of Transcendental Functions The following functions are continuous at all points of their domains. Trigonometric cos x sin x tan x cot x sec x csc x Inverse Trigonometric cos-1 x sin-1 x cot-1 x tan-1 x -1 csc -1 x sec x Exponential ex bx Logarithmic logb x ln x For each function listed in Theorem 2.15, we have lim f 1x2 = f 1a2, provided a is in xSa the domain of the function. This means that limits involving these functions may be evaluated by direct substitution at points in the domain. EXAMPLE 7 Limits involving transcendental functions Evaluate the following limits after determining the continuity of the functions involved. cos2 x - 1 4 a. lim b. lim 1 2 ln x + tan-1 x 2 x S 0 cos x - 1 xS1 SOLUTION ➤ Limits like the one in Example 7a are denoted 0/0 and are known as indeterminate forms, to be studied further in Section 4.7. a. Both cos2 x - 1 and cos x - 1 are continuous for all x by Theorems 2.9 and 2.15. However, the ratio of these functions is continuous only when cos x - 1 ≠ 0, which occurs when x is not an integer multiple of 2p. Note that both the numerator and denominator cos2 x - 1 of approach 0 as x S 0. To evaluate the limit, we factor and simplify: cos x - 1 lim xS0 1cos x - 121cos x + 12 cos2 x - 1 = lim = lim 1cos x + 12 cos x - 1 xS0 cos x - 1 xS0 (where cos x - 1 may be canceled because it is nonzero as x approaches 0). The limit on the right is now evaluated using direct substitution: lim 1cos x + 12 = cos 0 + 1 = 2. xS0 4 Show that f 1x2 = 2ln x is right-continuous at x = 1. xS1 xS1 does not exist. Related Exercises 51–56 ➤ QUICK CHECK 5 b. By Theorem 2.15, ln x is continuous on its domain 10, ∞2. However, ln x 7 0 only 4 when x 7 1, so Theorem 2.13 implies 2 ln x is continuous on 11, ∞2. At x = 1, 4 2ln x is right-continuous (Quick Check 5). The domain of tan-1 x is all real 4 numbers, and it is continuous on 1- ∞, ∞2. Therefore, f 1x2 = 2 ln x + tan-1 x is continuous on 31, ∞2. Because the domain of f does not include points with x 6 1, 4 4 lim- 1 2 ln x + tan-1 x 2 does not exist, which implies that lim 1 2 ln x + tan-1 x 2 We close this section with an important theorem that has both practical and theoretical uses. The Intermediate Value Theorem A common problem in mathematics is finding solutions to equations of the form f 1x2 = L. Before attempting to find values of x satisfying this equation, it is worthwhile to determine whether a solution exists. Copyright © 2014 Pearson Education, Inc. ➤ For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity 115 The existence of solutions is often established using a result known as the Intermediate Value Theorem. Given a function f and a constant L, we assume L lies between f 1a2 and f 1b2. The Intermediate Value Theorem says that if f is continuous on 3a, b4 , then the graph of f must cross the horizontal line y = L at least once (Figure 2.52). Although this theorem is easily illustrated, its proof goes beyond the scope of this text. Intermediate Value Theorem y y f (b) y f (x) f (a) y f (x) L L f (b) f (a) O a c b x O a c1 c2 c3 b x In (a, b), there is at least one number c such that f (c) L, where L is between f (a) and f (b). FIGURE 2.52 THEOREM 2.16 The Intermediate Value Theorem f is not continuous on [a, b]... f (b) Suppose f is continuous on the interval 3a, b4 and L is a number strictly between f 1a2 and f 1b2. Then there exists at least one number c in 1a, b2 satisfying f 1c2 = L. y ⫽ f (x) L The importance of continuity in Theorem 2.16 is illustrated in Figure 2.53, where we see a function f that is not continuous on 3a, b4 . For the value of L shown in the figure, there is no value of c in 1a, b2 satisfying f 1c2 = L. The next example illustrates a practical application of the Intermediate Value Theorem. f (a) O a x b EXAMPLE 8 Finding an interest rate Suppose you invest $1000 in a special 5-year ... and there is no number c in (a, b) such that f (c) ⫽ L. savings account with a fixed annual interest rate r, with monthly compounding. The r 60 amount of money A in the account after 5 years (60 months) is A1r2 = 1000a1 + b . 12 Your goal is to have $1400 in the account after 5 years. FIGURE 2.53 QUICK CHECK 6 Does the equation f 1x2 = x 3 + x + 1 = 0 have a solution on the interval 3 -1, 14 ? Explain. a. Use the Intermediate Value Theorem to show there is a value of r in (0, 0.08)—that is, an interest rate between 0% and 8%—for which A1r2 = 1400. ➤ b. Use a graphing utility to illustrate your explanation in part (a), and then estimate the interest rate required to reach your goal. Amount of money (dollars) after 5 years y SOLUTION y ⫽ A(r) 2000 y ⫽ 1400 Interest rate that yields $1400 after 5 years 500 r 60 b is continuous a. As a polynomial in r (of degree 60), A1r2 = 1000a1 + 12 for all r. Evaluating A1r2 at the endpoints of the interval 30, 0.084 , we have A102 = 1000 and A10.082 ≈ 1489.85. Therefore, A102 6 1400 6 A10.082, 0 0.02 0.0675 Interest rate FIGURE 2.54 0.10 r and it follows, by the Intermediate Value Theorem, that there is a value of r in 10, 0.082 for which A1r2 = 1400. b. The graphs of y = A1r2 and the horizontal line y = 1400 are shown in Figure 2.54; it is evident that they intersect between r = 0 and r = 0.08. Solving A1r2 = 1400 algebraically or using a root finder reveals that the curve and line intersect at r ≈ 0.0675. Therefore, an interest rate of approximately 6.75% is required for the investment to be worth $1400 after 5 years. Related Exercises 57–64 Copyright © 2014 Pearson Education, Inc. ➤ y For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 116 Limits SECTION 2.6 EXERCISES 13–20. Continuity at a point Determine whether the following functions are continuous at a. Use the continuity checklist to justify your answer. Review Questions 1. Which of the following functions are continuous for all values in their domain? Justify your answers. a. a1t2 = altitude of a skydiver t seconds after jumping from a plane b. n1t2 = number of quarters needed to park in a metered parking space for t minutes c. T1t2 = temperature t minutes after midnight in Chicago on January 1 d. p1t2 = number of points scored by a basketball player after t minutes of a basketball game 2. What does it mean for a function to be continuous on an interval? 4. We informally describe a function f to be continuous at a if its graph contains no holes or breaks at a. Explain why this is not an adequate definition of continuity. 5. 2x 2 + 3x + 1 ; a = 5 x 2 + 5x 14. f 1x2 = 2x 2 + 3x + 1 ; a = -5 x 2 + 5x 15. f 1x2 = 1x - 2; a = 1 Give the three conditions that must be satisfied by a function to be continuous at a point. 3. 13. f 1x2 = 16. g1x2 = 1 ; a = 3 x - 3 x2 - 1 17. f 1x2 = c x - 1 3 if x ≠ 1 ; a = 1 if x = 1 Complete the following sentences. x 2 - 4x + 3 x - 3 18. f 1x2 = c 2 a. A function is continuous from the left at a if __________. b. A function is continuous from the right at a if __________. 19. f 1x2 = if x ≠ 3 ; a = 3 if x = 3 5x - 2 ; a = 4 x - 9x + 20 2 6. Describe the points (if any) at which a rational function fails to be continuous. 7. What is the domain of f 1x2 = e x >x and where is f continuous? x2 + x 20. f 1x2 = c x + 1 2 8. Explain in words and pictures what the Intermediate Value Theorem says. 21–26. Continuity on intervals Use Theorem 2.10 to determine the intervals on which the following functions are continuous. Basic Skills 9–12. Discontinuities from a graph Determine the points at which the following functions f have discontinuities. For each point, state the conditions in the continuity checklist that are violated. 9. 10. y 4 3 3 2 2 1 1 0 11. 4 y ⫽ f (x) 1 2 3 4 5 0 x y 12. 5 4 3 3 2 2 1 1 1 2 3 4 1 2 3 5 x 4 5 0 22. g1x2 = 3x 2 - 6x + 7 x2 + x + 1 23. f 1x2 = x 5 + 6x + 17 x2 - 9 24. s1x2 = x 2 - 4x + 3 x2 - 1 25. f 1x2 = 1 x2 - 4 26. f 1t2 = t + 2 t2 - 4 27–30. Limits of compositions Evaluate the following limits and justify your answer. 4 3 27. lim 1x 8 - 3x 6 - 1240 b 28. lim a 5 xS0 x S 2 2x - 4x 2 - 50 x y 4 0 y ⫽ f (x) 29. lim a xS1 x + 5 4 b x + 2 30. lim a xS∞ 2x + 1 3 b x 31–34. Limits of composite functions Evaluate the following limits and justify your answer. 5 y ⫽ f (x) 21. p1x2 = 4x 5 - 3x 2 + 1 y 5 5 if x ≠ -1 ; a = -1 if x = - 1 y ⫽ f (x) 1 2 3 4 5 x 3 - 2x 2 - 8x xS4 B x - 4 x 31. lim 32. lim tan sin x 33. lim ln a2 b x xS0 34. lim a tS4 xS0 t - 4 2t - 2 x 216x + 1 - 1 b 1>3 35–38. Intervals of continuity Determine the intervals of continuity for the following functions. 35. The graph of Exercise 9 Copyright © 2014 Pearson Education, Inc. 36. The graph of Exercise 10 For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity 37. The graph of Exercise 11 38. The graph of Exercise 12 b. Use a graph to illustrate your explanation in part (a); then approximate the interest rate required to reach your goal. 39. Intervals of continuity Let T x 2 + 3x if x Ú 1 f 1x2 = e 2x if x 6 1. a. Use the continuity checklist to show that f is not continuous at 0. b. Is f continuous from the left or right at 0? c. State the interval(s) of continuity. 42. g1x2 = 2x 4 - 1 3 2 43. f 1x2 = 2 x - 2x - 3 44. f 1t2 = 1t 2 - 123>2 45. f 1x2 = 12x - 322>3 46. f 1z2 = 1z - 123>4 49. lim xS3 1 2x 2 48. + 72 lim xS - 1 50. lim tS2 1x 2 52. f 1x2 = e 1x; lim f 1x2; xS4 53. f 1x2 = 1 + sin x ; cos x 54. f 1x2 = ln x ; sin-1 x x S 1- ex 55. f 1x2 = ; 1 - ex x S 0- 56. f 1x2 = T lim f 1x2; x S p>4 - 4 + 2x - 9 2 3 2 e 2x - 1 ; ex - 1 61. x 3 - 5x 2 + 2x = - 1; 1-1, 52 62. -x 5 - 4x 2 + 21x + 5 = 0; 10, 32 1 + 2t 2 + 5 64. x ln x - 1 = 0; 11, e2 lim f 1x2 lim f 1x2 x S 4p>3 lim f 1x2 lim f 1x2; 60. 2x 4 + 25x 3 + 10 = 5; 10, 12 63. x + e x = 0; 1-1, 02 x S 2p- f 1x2; a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval. t2 + 5 lim f 1x2 lim 59–64. Applying the Intermediate Value Theorem 59. 2x 3 + x - 2 = 0; 1-1, 12 x S 0+ x S p>2- T c. Illustrate your answers with an appropriate graph. 51–56. Continuity and limits with transcendental functions Determine the interval(s) on which the following functions are continuous; then evaluate the given limits. 51. f 1x2 = csc x; , b. Use a graphing utility to find all the solutions to the equation on the given interval. 47–50. Limits with roots Determine the following limits and justify your answers. 4x + 10 47. lim x S 2 A 2x - 2 1 - 11 + r>122-360 a. Use the Intermediate Value Theorem to show there is a value of r in (0.06, 0.08)—an interest rate between 6% and 8%—that allows you to make monthly payments of $1000 per month. b. Use a graph to illustrate your explanation to part (a). Then determine the interest rate you need for monthly payments of $1000. x 3 + 4x + 1 if x … 0 if x 7 0. 2x 3 41. f 1x2 = 22x 2 - 16 150,0001r>122 where r is the annual interest rate. Suppose banks are currently offering interest rates between 6% and 8%. 40. Intervals of continuity Let 41–46. Functions with roots Determine the interval(s) on which the following functions are continuous. Be sure to consider right- and left-continuity at the endpoints. 58. Intermediate Value Theorem and mortgage payments You are shopping for a $150,000, 30-year (360-month) loan to buy a house. The monthly payment is m1r2 = a. Use the continuity checklist to show that f is not continuous at 1. b. Is f continuous from the left or right at 1? c. State the interval(s) of continuity. f 1x2 = e 117 Further Explorations 65. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If a function is left-continuous and right-continuous at a, then it is continuous at a. b. If a function is continuous at a, then it is left-continuous and right-continuous at a. c. If a 6 b and f 1a2 … L … f 1b2, then there is some value of c in 1a, b2 for which f 1c2 = L. d. Suppose f is continuous on 3a, b4. Then there is a point c in 1a, b2 such that f 1c2 = 1 f 1a2 + f 1b22>2. 66. Continuity of the absolute value function Prove that the absolute value function x is continuous for all values of x. (Hint: Using the definition of the absolute value function, compute lim- x and lim+ x .) lim f 1x2 x S 0+ xS0 lim f 1x2 xS0 57. Intermediate Value Theorem and interest rates Suppose $5000 is invested in a savings account for 10 years (120 months), with an annual interest rate of r, compounded monthly. The amount of money in the account after 10 years is A1r2 = 500011 + r>122120. a. Use the Intermediate Value Theorem to show there is a value of r in (0, 0.08)—an interest rate between 0% and 8%—that allows you to reach your savings goal of $7000 in 10 years. xS0 67–70. Continuity of functions with absolute values Use the continuity of the absolute value function (Exercise 66) to determine the interval(s) on which the following functions are continuous. 67. f 1x2 = x 2 + 3x - 18 69. h1x2 = ` 1 ` 1x - 4 Copyright © 2014 Pearson Education, Inc. 68. g1x2 = ` x + 4 ` x2 - 4 70. h1x2 = x 2 + 2x + 5 + 1x For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 118 Limits 71–80. Miscellaneous limits Evaluate the following limits. cos2 x + 3 cos x + 2 cos x + 1 71. lim xSp sin x - 1 73. lim x S p>2 1sin x - 1 75. lim xS0 cos x - 1 sin2 x 72. lim x S 3p>2 sin2 x + 6 sin x + 5 sin2 x - 1 1 1 2 + sin u 2 74. lim uS0 sin u 76. lim+ xS0 -1 tan x x 78. lim cos t e 3t 79. lim- x ln x 80. lim+ x ln x xS1 tS ∞ xS0 T 1 - cos2 x sin x 77. lim xS∞ a. Determine the value of a for which g is continuous from the left at 1. b. Determine the value of a for which g is continuous from the right at 1. c. Is there a value of a for which g is continuous at 1? Explain. 86. Asymptotes of a function containing exponentials Let 2e x + 5e 3x f 1x2 = 2x . Evaluate lim- f 1x2, lim+ f 1x2, lim f 1x2, xS0 xS0 x S -∞ e - e 3x and lim f 1x2. Then give the horizontal and vertical asymptotes xS∞ of f. Plot f to verify your results. T 81. Pitfalls using technology The graph of the sawtooth function y = x - : x ; , where : x ; is the greatest integer function or floor function (Exercise 37, Section 2.2), was obtained using a graphing utility (see figure). Identify any inaccuracies appearing in the graph and then plot an accurate graph by hand. y ⫽ x ⫺ :x; 87. Asymptotes of a function containing exponentials Let 2e x + 10e - x . Evaluate lim f 1x2, lim f 1x2, and f 1x2 = ex + e - x xS0 x S -∞ lim f 1x2. Then give the horizontal and vertical asymptotes of f. xS∞ Plot f to verify your results. T 88–89. Applying the Intermediate Value Theorem Use the Intermediate Value Theorem to verify that the following equations have three solutions on the given interval. Use a graphing utility to find the approximate roots. 88. x 3 + 10x 2 - 100x + 50 = 0; 1-20, 102 1.5 89. 70x 3 - 87x 2 + 32x - 3 = 0; 10, 12 Applications ⫺2 90. Parking costs Determine the intervals of continuity for the parking cost function c introduced at the outset of this section (see figure). Consider 0 … t … 60. 2 y ⫺0.5 82. Pitfalls using technology Graph the function f 1x2 = a graphing window of 3- p, p4 * 30, 24 . sin x using x a. Sketch a copy of the graph obtained with your graphing device and describe any inaccuracies appearing in the graph. b. Sketch an accurate graph of the function. Is f continuous at 0? sin x c. What is the value of lim . S x 0 x 1.25 Cost (dollars) T 1.00 0.50 0.25 0 a. Sketch the graph of a function that is not continuous at 1, but is defined at 1. b. Sketch the graph of a function that is not continuous at 1, but has a limit at 1. 84. An unknown constant Determine the value of the constant a for which the function if x ≠ - 1 if x = - 1 is continuous at -1. 85. An unknown constant Let x2 + x g1x2 = W a 3x + 5 15 30 45 60 t Time (min) 83. Sketching functions x 2 + 3x + 2 x + 1 f 1x2 = c a y ⫽ c(t) 0.75 if x 6 1 if x = 1 if x 7 1. 91. Investment problem Assume you invest $250 at the end of each year for 10 years at an annual interest rate of r. The amount of money 250111 + r210 - 12 in your account after 10 years is A = . r Assume your goal is to have $3500 in your account after 10 years. a. Use the Intermediate Value Theorem to show that there is an interest rate r in the interval 10.01, 0.102—between 1% and 10%—that allows you to reach your financial goal. b. Use a calculator to estimate the interest rate required to reach your financial goal. 92. Applying the Intermediate Value Theorem Suppose you park your car at a trailhead in a national park and begin a 2-hr hike to a lake at 7 a.m. on a Friday morning. On Sunday morning, you leave the lake at 7 a.m. and start the 2-hr hike back to your car. Assume the lake is 3 mi from your car. Let f 1t2 be your distance from the car t hours after 7 a.m. on Friday morning and let g1t2 be your distance from the car t hours after 7 a.m. on Sunday morning. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.6 Continuity a. Evaluate f 102, f 122, g102, and g122. b. Let h1t2 = f 1t2 - g1t2. Find h102 and h122. c. Use the Intermediate Value Theorem to show that there is some point along the trail that you will pass at exactly the same time of morning on both days. 93. The monk and the mountain A monk set out from a monastery in the valley at dawn. He walked all day up a winding path, stopping for lunch and taking a nap along the way. At dusk, he arrived at a temple on the mountaintop. The next day, the monk made the return walk to the valley, leaving the temple at dawn, walking the same path for the entire day, and arriving at the monastery in the evening. Must there be one point along the path that the monk occupied at the same time of day on both the ascent and descent? (Hint: The question can be answered without the Intermediate Value Theorem.) (Source: Arthur Koestler, The Act of Creation.) 94. Does continuity of f imply continuity of f? Let 1 if x Ú 0 - 1 if x 6 0. Write a formula for g1x2 . Is g continuous at x = 0? Explain. Is g continuous at x = 0? Explain. For any function f, if f is continuous at a, does it necessarily follow that f is continuous at a? Explain. 95–96. Classifying discontinuities The discontinuities in graphs (a) and (b) are removable discontinuities because they disappear if we define or redefine f at a so that f 1a2 = lim f 1x2. The function in xSa graph (c) has a jump discontinuity because left and right limits exist at a but are unequal. The discontinuity in graph (d) is an infinite discontinuity because the function has a vertical asymptote at a. y y y ⫽ f (x) y a (a) T 100–101. Classifying discontinuities Classify the discontinuities in the following functions at the given points. See Exercises 95–96. 100. f 1x2 = O x - 2 x 3 - 4x 2 + 4x ; x = 0 and x = 1 x1x - 12 102. Continuity of composite functions Prove Theorem 2.11: If g is continuous at a and f is continuous at g1a2, then the composition f ∘ g is continuous at a. (Hint: Write the definition of continuity for f and g separately; then combine them to form the definition of continuity for f ∘ g.) 103. Continuity of compositions a. Find functions f and g such that each function is continuous at 0, but f ∘ g is not continuous at 0. b. Explain why examples satisfying part (a) do not contradict Theorem 2.11. 104. Violation of the Intermediate Value Theorem? Let x f 1x2 = . Then f 1-22 = - 1 and f 122 = 1. Therefore, x f 1-22 6 0 6 f 122, but there is no value of c between - 2 and 2 for which f 1c2 = 0. Does this fact violate the Intermediate Value Theorem? Explain. xS0 x a (b) xS0 y ⫽ f (x) y ⫽ f (x) a (c) x Infinite discontinuity O a (d) x 95. Is the discontinuity at a in graph (c) removable? Explain. 96. Is the discontinuity at a in graph (d) removable? Explain. 97–98. Removable discontinuities Show that the following functions have a removable discontinuity at the given point. See Exercises 95–96. 97. f 1x2 = xSa QUICK CHECK ANSWERS 1. t = 15, 30, 45 2. Both expressions have a value of 5, showing that lim f 1g1x22 = f 1lim g1x22. 3. Fill xSa in the endpoints. O xSa thereby establishing that sin x is continuous for all x. (Hint: Let h = x - a so that x = a + h and note that h S 0 as x S a.) b. Use the identity cos 1a + h2 = cos a cos h - sin a sin h with the fact that lim cos x = 1 to prove that lim cos x = cos a. y Jump discontinuity ; x = 2 a. Use the identity sin 1a + h2 = sin a cos h + cos a sin h with the fact that lim sin x = 0 to prove that lim sin x = sin a, Removable discontinuity x x - 2 105. Continuity of sin x and cos x y ⫽ f (x) Removable discontinuity O a. Does the function f 1x2 = x sin 11>x2 have a removable discontinuity at x = 0? b. Does the function g1x2 = sin 11>x2 have a removable discontinuity at x = 0? 4 xSa 4. 30, ∞2; 1- ∞, ∞2 5. Note that 4 lim 2ln x = 2 lim ln x = 0 and f 112 = 2 ln 1 = 0. xS1 + 4 x S 1+ Because the limit from the right and the value of the function at x = 1 are equal, the function is right-continuous at x = 1. 6. The equation has a solution on the interval 3 -1, 14 because f is continuous on 3 -1, 14 and f 1-12 6 0 6 f 112. x 2 - 7x + 10 ; x = 2 x - 2 Copyright © 2014 Pearson Education, Inc. ➤ g1x2 = e if x ≠ 1 ; x = 1 if x = 1 99. Do removable discontinuities exist? Refer to Exercises 95–96. 101. h1x2 = Additional Exercises a. b. c. d. x2 - 1 98. g1x2 = c 1 - x 3 119 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 120 Limits 2.7 Precise Definitions of Limits The limit definitions already encountered in this chapter are adequate for most elementary limits. However, some of the terminology used, such as sufficiently close and arbitrarily large, needs clarification. The goal of this section is to give limits a solid mathematical foundation by transforming the previous limit definitions into precise mathematical statements. Moving Toward a Precise Definition Assume the function f is defined for all x near a, except possibly at a. Recall that lim f 1x2 = L means that f 1x2 is arbitrarily close to L for all x sufficiently close (but not xSa equal) to a. This limit definition is made precise by observing that the distance between f 1x2 and L is f 1x2 - L and that the distance between x and a is x - a . Therefore, we write lim f 1x2 = L if we can make f 1x2 - L arbitrarily small for any x, distinct from a, xSa with x - a sufficiently small. For instance, if we want f 1x2 - L to be less than 0.1, then we must find a number d 7 0 such that ➤ The phrase for all x near a means for all f 1x2 - L 6 0.1 whenever x in an open interval containing a. x - a 6 d and x ≠ a. If, instead, we want f 1x2 - L to be less than 0.001, then we must find another number d 7 0 such that ➤ The Greek letters d (delta) and e (epsilon) represent small positive numbers when discussing limits. f 1x2 - L 6 0.001 whenever 0 6 x - a 6 d. For the limit to exist, it must be true that for any e 7 0, we can always find a d 7 0 such that ➤ The two conditions x - a 6 d and x ≠ a are written concisely as 0 6 x - a 6 d. f 1x2 - L 6 e whenever 0 6 x - a 6 d. y EXAMPLE 1 Determining values of D from a graph Figure 2.55 shows the graph of 7 a linear function f with lim f 1x2 = 5. For each value of e 7 0, determine a value of xS3 d 7 0 satisfying the statement y f (x) 5 f 1x2 - 5 6 e whenever 0 6 x - 3 6 d. lim f (x) 5 ᠬ x 3 a. e = 1 3 b. e = 1 0 1 2 SOLUTION 1 3 5 7 FIGURE 2.55 ➤ The founders of calculus, Isaac Newton (1642–1727) and Gottfried Leibniz (1646–1716), developed the core ideas of calculus without using a precise definition of a limit. It was not until the 19th century that a rigorous definition was introduced by Louis Cauchy (1789–1857) and later refined by Karl Weierstrass (1815–1897). x a. With e = 1, we want f 1x2 to be less than 1 unit from 5, which means f 1x2 is between 4 and 6. To determine a corresponding value of d, draw the horizontal lines y = 4 and y = 6 (Figure 2.56a). Then sketch vertical lines passing through the points where the horizontal lines and the graph of f intersect (Figure 2.56b). We see that the vertical lines intersect the x-axis at x = 1 and x = 5. Note that f 1x2 is less than 1 unit from 5 on the y-axis if x is within 2 units of 3 on the x-axis. So for e = 1, we let d = 2 or any smaller positive value. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.7 Precise Definitions of Limits f (x) 5 1 y y 6 6 5 5 4 4 0 1 3 5 7 x 121 Values of x such that f (x) 5 1 f (x) 5 1 0 1 3 5 7 x 0 x 3 2 (b) (a) FIGURE 2.56 ➤ Once an acceptable value of d is found satisfying the statement f 1x2 - L 6 e whenever 0 6 x - a 6 d, b. With e = 12, we want f 1x2 to lie within a half-unit of 5 or, equivalently, f 1x2 must lie between 4.5 and 5.5. Proceeding as in part (a), we see that f 1x2 is within a half-unit of 5 on the y-axis (Figure 2.57a) if x is less than 1 unit from 3 (Figure 2.57b). So for e = 12, we let d = 1 or any smaller positive number. any smaller positive value of d also works. y 5.5 5 4.5 5.5 5 4.5 0 2 3 x 4 FIGURE 2.57 5Ω 3 x 4 The idea of a limit, as illustrated in Example 1, may be described in terms of a contest between two people named Epp and Del. First, Epp picks a particular number e 7 0; then he challenges Del to find a corresponding value of d 7 0 such that f 1x2 - 5 6 e whenever 0 6 x - 3 6 d. 5 f (x) 5 Ω 4 0 2 Related Exercises 9–12 Values of x such that f (x) 5 Ω 6 5Ω f (x) 5 q 0 x 3 1 (b) (a) y 0 Values of x such that f (x) 5 q ➤ f (x) 5 q y 2 3~ 3 3~ 0 x 3 ~ FIGURE 2.58 4 x (1) To illustrate, suppose Epp chooses e = 1. From Example 1, we know that Del will satisfy (1) by choosing 0 6 d … 2. If Epp chooses e = 12, then (by Example 1) Del responds by letting 0 6 d … 1. If Epp lets e = 18, then Del chooses 0 6 d … 14 (Figure 2.58). In fact, there is a pattern: For any e 7 0 that Epp chooses, no matter how small, Del will satisfy (1) by choosing a positive value of d satisfying 0 6 d … 2e. Del has discovered a mathematical relationship: If 0 6 d … 2e and 0 6 x - 3 6 d, then f 1x2 - 5 6 e, for any e 7 0. This conversation illustrates the general procedure for proving that lim f 1x2 = L. xSa Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Limits QUICK CHECK 1 In Example 1, find a positive number d satisfying the statement f 1x2 - 5 6 1 100 whenever 0 6 x - 3 6 d. ➤ Chapter 2 r 122 A Precise Definition Example 1 dealt with a linear function, but it points the way to a precise definition of a limit for any function. As shown in Figure 2.59, lim f 1x2 = L means that for any positive xSa number e, there is another positive number d such that f 1x2 - L 6 e whenever 0 6 x - a 6 d. In all limit proofs, the goal is to find a relationship between e and d that gives an admissible value of d, in terms of e only. This relationship must work for any positive value of e. lim f (x) L ᠬ x a y L⑀ f (x) L ... then f (x) L ⑀. L⑀ y f (x) a␦ x a O a␦ x If 0 x a ␦... FIGURE 2.59 DEFINITION Limit of a Function ➤ The value of d in the precise definition of a limit depends only on e. Assume that f 1x2 exists for all x in some open interval containing a, except possibly at a. We say that the limit of f 1x2 as x approaches a is L, written lim f 1x2 = L, xSa ➤ Definitions of the one-sided limits lim f 1x2 = L and lim- f 1x2 = L are x S a+ xSa if for any number e 7 0 there is a corresponding number d 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d. discussed in Exercises 39–43. EXAMPLE 2 Finding D for a given E using a graphing utility Let f 1x2 = x 3 - 6x 2 + 12x - 5 and demonstrate that lim f 1x2 = 3 as follows. xS2 For the given values of e, use a graphing utility to find a value of d 7 0 such that f 1x2 - 3 6 e whenever 0 6 x - 2 6 d. a. e = 1 b. e = 1 2 SOLUTION a. The condition f 1x2 - 3 6 e = 1 implies that f 1x2 lies between 2 and 4. Using a graphing utility, we graph f and the lines y = 2 and y = 4 (Figure 2.60). These lines intersect the graph of f at x = 1 and at x = 3. We now sketch the vertical lines Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.7 Precise Definitions of Limits 123 x = 1 and x = 3 and observe that f 1x2 is within 1 unit of 3 whenever x is within 1 unit of 2 on the x-axis (Figure 2.60). Therefore, with e = 1, we can choose any d with 0 6 d … 1. b. The condition f 1x2 - 3 6 e = 12 implies that f 1x2 lies between 2.5 and 3.5 on the y-axis. We now find that the lines y = 2.5 and y = 3.5 intersect the graph of f at x ≈ 1.21 and x ≈ 2.79 (Figure 2.61). Observe that if x is less than 0.79 units from 2 on the x-axis, then f 1x2 is less than a half-unit from 3 on the y-axis. Therefore, with e = 12 we can choose any d with 0 6 d … 0.79. y y 5 y f (x) y f (x) 4 3.5 f (x) 3 1 3 f (x) 3 q 3 2.5 2 1 0 1 2 3 x 0 0 x 2 1 FIGURE 2.60 1.21 2 2.79 4 0 x 2 0.79 FIGURE 2.61 Related Exercises 13–14 For the function f given in Example 2, estimate a value of d 7 0 satisfying f 1x2 - 3 6 0.25 whenever 0 6 x - 2 6 d. ➤ This procedure could be repeated for smaller and smaller values of e 7 0. For each value of e, there exists a corresponding value of d, proving that the limit exists. QUICK CHECK 2 x The inequality 0 6 x - a 6 d means that x lies between a - d and a + d with x ≠ a. We say that the interval 1a - d, a + d2 is symmetric about a because a is the midpoint of the interval. Symmetric intervals are convenient, but Example 3 demonstrates that we don’t always get symmetric intervals without a bit of extra work. ➤ EXAMPLE 3 Finding a symmetric interval Figure 2.62 shows the graph of g with y lim g1x2 = 3. For each value of e, find the corresponding values of d 7 0 that satisfy the condition xS2 6 g1x2 - 3 6 e whenever 0 6 x - 2 6 d. y g(x) a. e = 2 b. e = 1 c. For any given value of e, make a conjecture about the corresponding values of d that satisfy the limit condition. 3 2 1 SOLUTION 0 1 FIGURE 2.62 2 6 x a. With e = 2, we need a value of d 7 0 such that g1x2 is within 2 units of 3, which means between 1 and 5, whenever x is less than d units from 2. The horizontal lines y = 1 and y = 5 intersect the graph of g at x = 1 and x = 6. Therefore, g1x2 - 3 6 2 if x lies in the interval 11, 62 with x ≠ 2 (Figure 2.63a). However, we want x to lie in an interval that is symmetric about 2. We can guarantee that g1x2 - 3 6 2 only if x is less than 1 unit away from 2, on either side of 2 (Figure 2.63b). Therefore, with e = 2, we take d = 1 or any smaller positive number. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 124 Chapter 2 r Limits g(x) 3 2 y y 5 5 y g(x) y g(x) 3 3 1 1 0 1 2 3 6 x 0 1 2 3 6 x Symmetric interval 0 x 2 1 that guarantees g(x) 3 2 Values of x such that g(x) 3 2 (a) (b) FIGURE 2.63 b. With e = 1, g1x2 must lie between 2 and 4 (Figure 2.64a). This implies that x must be within a half-unit to the left of 2 and within 2 units to the right of 2. Therefore, g1x2 - 3 6 1 provided x lies in the interval 11.5, 42. To obtain a symmetric interval about 2, we take d = 12 or any smaller positive number. Then we are guaranteed that g1x2 - 3 6 1 when 0 6 x - 2 6 12 (Figure 2.64b). y y 4 g(x) 3 1 4 y g(x) 3 3 2 2 0 1.5 4 x 0 y g(x) 1.5 2.5 4 x Symmetric interval 0 x 2 q that guarantees g(x) 3 1 Values of x such that g(x) 3 1 (b) (a) FIGURE 2.64 c. From parts (a) and (b), it appears that if we choose d … e>2, the limit condition is satisfied for any e 7 0. Limit Proofs We use the following two-step process to prove that lim f 1x2 = L. xSa Steps for proving that lim f 1x2 = L xSa ➤ The first step of the limit-proving process is the preliminary work of finding a candidate for d. The second step verifies that the d found in the first step actually works. 1. Find D. Let e be an arbitrary positive number. Use the inequality f 1x2 - L 6 e to find a condition of the form x - a 6 d, where d depends only on the value of e. 2. Write a proof. For any e 7 0, assume 0 6 x - a 6 d and use the relationship between e and d found in Step 1 to prove that f 1x2 - L 6 e. Copyright © 2014 Pearson Education, Inc. ➤ Related Exercises 15–18 For Use Only in 2013 – 2014 Pilot Program 2.7 Precise Definitions of Limits 125 EXAMPLE 4 Limit of a linear function Prove that lim 14x - 152 = 1 using the xS4 precise definition of a limit. SOLUTION Step 1: Find d. In this case, a = 4 and L = 1. Assuming e 7 0 is given, we use 14x - 152 - 1 6 e to find an inequality of the form x - 4 6 d. If 14x - 152 - 1 6 e, then 4x - 16 6 e 4 x - 4 6 e Factor 4x - 16. e x - 4 6 . Divide by 4 and identify d = e>4. 4 We have shown that 14x - 152 - 1 6 e implies x - 4 6 e>4. Therefore, a plausible relationship between d and e is d = e>4. We now write the actual proof. Step 2: Write a proof. Let e 7 0 be given and assume 0 6 x - 4 6 d where d = e>4. The aim is to show that 14x - 152 - 1 6 e for all x such that 0 6 x - 4 6 d. We simplify 14x - 152 - 1 and isolate the x - 4 term: 14x - 152 - 1 = 4x - 16 = 4 x - 4 (+)+* less than d = e>4 e 6 4a b = e. 4 We have shown that for any e 7 0, f 1x2 - L = 14x - 152 - 1 6 e whenever 0 6 x - 4 6 d, xS4 Related Exercises 19–24 ➤ provided 0 6 d … e>4. Therefore, lim 14x - 152 = 1. Justifying Limit Laws The precise definition of a limit is used to prove the limit laws in Theorem 2.3. Essential in several of these proofs is the triangle inequality, which states that x + y … x + y , for all real numbers x and y. EXAMPLE 5 Proof of Limit Law 1 Prove that if lim f 1x2 and lim g1x2 exist, then xSa xSa lim 3 f 1x2 + g1x24 = lim f 1x2 + lim g1x2. xSa ➤ Because lim f 1x2 exists, if there exists a xSa d 7 0 for any given e 7 0, then there also exists a d 7 0 for any given 2e . xSa xSa SOLUTION Assume that e 7 0 is given. Let lim f 1x2 = L, which implies that there xSa exists a d1 7 0 such that f 1x2 - L 6 e 2 whenever 0 6 x - a 6 d1. Similarly, let lim g1x2 = M, which implies there exists a d2 7 0 such that xSa g1x2 - M 6 e 2 whenever 0 6 x - a 6 d2. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Limits ➤ The minimum value of a and b is denoted min 5a, b6. If x = min 5a, b6, then x is the smaller of a and b. If a = b, then x equals the common value of a and b. In either case, x … a and x … b. Let d = min 5d1, d26 and suppose 0 6 x - a 6 d. Because d … d1, it follows that 0 6 x - a 6 d1 and f 1x2 - L 6 e>2. Similarly, because d … d2, it follows that 0 6 x - a 6 d2 and g1x2 - M 6 e>2. Therefore, 0 1 f 1x2 + g1x22 - 1L + M2 0 = 0 1 f 1x2 - L2 + 1g1x2 - M2 0 Rearrange terms. … 0 f 1x2 - L 0 + 0 g1x2 - M 0 Triangle inequality. 6 ➤ Proofs of other limit laws are outlined in Exercises 25 and 26. We have shown that given any e 7 0, if 0 6 x - a 6 d, then 0 1 f 1x2 + g1x22 - 1L + M2 0 6 e, which implies that lim 3 f 1x2 + g1x24 = xSa L + M = lim f 1x2 + lim g1x2. Related Exercises 25–28 xSa ➤ Notice that for infinite limits, N plays the role that e plays for regular limits. It sets a tolerance or bound for the function values f 1x2. e e + = e. 2 2 xSa ➤ Chapter 2 r 126 Infinite Limits In Section 2.4, we stated that lim f 1x2 = ∞ if f 1x2 grows arbitrarily large as x approaches xSa a. More precisely, this means that for any positive number N (no matter how large), f 1x2 is larger than N if x is sufficiently close to a but not equal to a. DEFINITION Two-Sided Infinite Limit The infinite limit lim f 1x2 = ∞ means that for any positive number N, there exists a xSa corresponding d 7 0 such that f 1x2 7 N whenever 0 6 x - a 6 d. As shown in Figure 2.65, to prove that lim f 1x2 = ∞, we let N represent any positive xSa number. Then we find a value of d 7 0, depending only on N, such that f 1x2 7 N whenever 0 6 x - a 6 d. This process is similar to the two-step process for finite limits. ➤ Precise definitions for lim f 1x2 = - ∞, y xSa lim+ f 1x2 = - ∞, lim+ f 1x2 = ∞, xSa xSa lim f 1x2 = - ∞, and lim- f 1x2 = ∞ x S a- xSa are given in Exercises 45–49. f (x) f (x) N N O a␦ x a a␦ 0 x a Values of x such that f (x) N FIGURE 2.65 Copyright © 2014 Pearson Education, Inc. x For Use Only in 2013 – 2014 Pilot Program 2.7 Precise Definitions of Limits 127 Steps for proving that lim f 1x2 = H xSa 1. Find D. Let N be an arbitrary positive number. Use the statement f 1x2 7 N to find an inequality of the form x - a 6 d, where d depends only on N. 2. Write a proof. For any N 7 0, assume 0 6 x - a 6 d and use the relationship between N and d found in Step 1 to prove that f 1x2 7 N. EXAMPLE 6 An Infinite Limit Proof Let f 1x2 = lim f 1x2 = ∞. 1 . Prove that 1x - 222 xS2 SOLUTION 1 7 N to find d, 1x - 222 where d depends only on N. Taking reciprocals of this inequality, it follows that Step 1: Find d 7 0. Assuming N 7 0, we use the inequality 1 N 1 x - 2 6 . 1N 1x - 222 6 ➤ Recall that 2x 2 = x . Take the square root of both sides. 1 1 has the form x - 2 6 d if we let d = . 1N 1N We now write a proof based on this relationship between d and N. The inequality x - 2 6 1 Step 2: Write a proof. Suppose N 7 0 is given. Let d = and assume 1N 1 0 6 x - 2 6 d = . Squaring both sides of the inequality 1N 1 x - 2 6 and taking reciprocals, we have 1N 1 N Square both sides. 1 7 N. Take reciprocals of both sides. 1x - 222 1 We see that for any positive N, if 0 6 x - 2 6 d = , then 1N 1 1 f 1x2 = 7 N. It follows that lim = ∞. Note that 2 S x 2 1x - 22 1x - 222 1 because d = , d decreases as N increases. 1N QUICK CHECK 3 In Example 6, if N is increased by a factor of 100, how must d change? ➤ Related Exercises 29–32 ➤ 1x - 222 6 Limits at Infinity Precise definitions can also be written for the limits at infinity lim f 1x2 = L and S lim f 1x2 = L. For discussion and examples, see Exercises 50 and 51.x ∞ x S -∞ Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 128 Chapter 2 r Limits SECTION 2.7 EXERCISES 11. Determining values of D from a graph The function f in the figure satisfies lim f 1x2 = 6. Determine the largest value of d 7 0 Review Questions 1. Suppose x lies in the interval 11, 32 with x ≠ 2. Find the smallest positive value of d such that the inequality 0 6 x - 2 6 d is true. 2. Suppose f 1x2 lies in the interval 12, 62. What is the smallest value of e such that f 1x2 - 4 6 e? 3. Which one of the following intervals is not symmetric about x = 5? a. 11, 92 b. 14, 62 c. 13, 82 xS3 satisfying each statement. a. If 0 6 x - 3 6 d, then f 1x2 - 6 6 3. b. If 0 6 x - 3 6 d, then f 1x2 - 6 6 1. y 9 d. 14.5, 5.52 4. Does the set 5x: 0 6 x - a 6 d6 include the point x = a? Explain. 5. State the precise definition of lim f 1x2 = L. 6. Interpret f 1x2 - L 6 e in words. 7. Suppose f 1x2 - 5 6 0.1 whenever 0 6 x 6 5. Find all values of d 7 0 such that f 1x2 - 5 6 0.1 whenever 0 6 x - 2 6 d. 8. Give the definition of lim f 1x2 = ∞ and interpret it using xSa pictures. y ⫽ f (x) 6 xSa 1 0 Determining values of D from a graph The function f in the figure satisfies lim f 1x2 = 5. Determine the largest value xS4 of d 7 0 satisfying each statement. 6 d, 6 2. 6 d, 6 1. x 6 d 7 0 satisfying each statement. xS2 a. If 0 6 x - 2 then f 1x2 - 5 b. If 0 6 x - 2 then f 1x2 - 5 3 12. Determining values of D from a graph The function f in the figure satisfies lim f 1x2 = 5. Determine the largest value of Basic Skills 9. 1 a. If 0 6 x - 4 6 d, then f 1x2 - 5 6 1. b. If 0 6 x - 4 6 d, then f 1x2 - 5 6 0.5. y y 8 8 y ⫽ f (x) 5 5 y ⫽ f (x) 1 0 1 1 x 2 0 10. Determining values of D from a graph The function f in the figure satisfies lim f 1x2 = 4. Determine the largest value xS2 T of d 7 0 satisfying each statement. a. If 0 6 x - 2 then f 1x2 - 4 b. If 0 6 x - 2 then f 1x2 - 4 6 d, 6 1. 6 d, 6 1>2. 1 4 8 x 13. Finding D for a given E using a graph Let f 1x2 = x 3 + 3 and note that lim f 1x2 = 3. For each value of e, use a graphing utility xS0 y to find a value of d 7 0 such that f 1x2 - 3 6 e whenever 0 6 x - 0 6 d. Sketch graphs illustrating your work. 8 a. e = 1 T y ⫽ f (x) 14. Finding D for a given E using a graph Let g1x2 = 2x 3 - 12x 2 + 26x + 4 and note that lim g1x2 = 24. xS2 For each value of e, use a graphing utility to find a value of d 7 0 such that g1x2 - 24 6 e whenever 0 6 x - 2 6 d. Sketch graphs illustrating your work. 4 a. e = 1 1 0 b. e = 0.5 1 2 4 x b. e = 0.5 15. Finding a symmetric interval The function f in the figure satisfies lim f 1x2 = 3. For each value of e, find a value of d 7 0 xS2 such that f 1x2 - 3 6 e whenever 0 6 x - 2 6 d. a. e = 1 Copyright © 2014 Pearson Education, Inc. b. e = 1 2 (2) For Use Only in 2013 – 2014 Pilot Program 2.7 Precise Definitions of Limits c. For any e 7 0, make a conjecture about the corresponding value of d satisfying (2). 129 x 2 - 7x + 12 = -1 xS3 x - 3 22. lim y 23. lim x 2 = 0 (Hint: Use the identity 2x 2 = x .) 6 24. lim 1x - 322 = 0 (Hint: Use the identity 2x 2 = x .) xS0 xS3 25. Proof of Limit Law 2 Suppose lim f 1x2 = L and lim g1x2 = M. y ⫽ f (x) xSa xSa Prove that lim 3f 1x2 - g1x24 = L - M. xSa 3 26. Proof of Limit Law 3 Suppose lim f 1x2 = L. Prove that xSa lim 3cf 1x24 = cL, where c is a constant. 1 xSa 0 1 2 27. Limit of a constant function and f 1x2 = x Give proofs of the following theorems. x 6 16. Finding a symmetric interval The function f in the figure satisfies lim f 1x2 = 5. For each value of e, find a value of d 7 0 xS4 such that f 1x2 - 5 6 e whenever 0 6 x - 4 6 d. (3) a. e = 2 b. e = 1 c. For any e 7 0, make a conjecture about the corresponding value of d satisfying (3). y a. lim c = c for any constant c b. lim x = a for any constant a xSa xSa 28. Continuity of linear functions Prove Theorem 2.2: If f 1x2 = mx + b, then lim f 1x2 = ma + b for constants m and b. (Hint: xSa For a given e 7 0, let d = e> m .) Explain why this result implies that linear functions are continuous. 29–32. Limit proofs for infinite limits Use the precise definition of infinite limits to prove the following limits. 7 29. lim 6 xS4 1 = ∞ 1x - 422 5 31. lim a 4 xS0 1 + 1b = ∞ x2 30. lim 1 = ∞ 1x + 124 32. lim a 1 - sin xb = ∞ x4 x S -1 xS0 3 2 Further Explorations 1 33. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume a and L are finite numbers and assume lim f 1x2 = L. 0 T 1 2 3 4 5 6 7 x 2x 2 - 2 17. Finding a symmetric interval Let f 1x2 = and note x - 1 that lim f 1x2 = 4. For each value of e, use a graphing utility xS1 to find a value of d 7 0 such that f 1x2 - 4 6 e whenever 0 6 x - 1 6 d. a. e = 2 b. e = 1 c. For any e 7 0, make a conjecture about the value of d that satisfies the preceding inequality. x + 1 if x … 3 1 if x 7 3 2x + 2 and note that lim f 1x2 = 2. For each value of e, use a graphing 1 T 18. Finding a symmetric interval Let f 1x2 = b 31 xS3 utility to find a value of d 7 0 such that f 1x2 - 2 6 e whenever 0 6 x - 3 6 d. b. e = 14 a. e = 12 c. For any e 7 0, make a conjecture about the value of d that satisfies the preceding inequality. 19–24. Limit proofs Use the precise definition of a limit to prove the following limits. 19. lim 18x + 52 = 13 xS1 20. lim 1- 2x + 82 = 2 xS3 21. lim xSa we can always find an e 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d. c. The limit lim f 1x2 = L means that for any arbitrary e 7 0, xSa we can always find a d 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d. d. If x - a 6 d, then a - d 6 x 6 a + d. 34. Finding D algebraically Let f 1x2 = x 2 - 2x + 3. a. For e = 0.25, find a corresponding value of d 7 0 satisfying the statement f 1x2 - 2 6 e whenever 0 6 x - 1 6 d. b. Verify that lim f 1x2 = 2 as follows. For any e 7 0, find a xS1 corresponding value of d 7 0 satisfying the statement f 1x2 - 2 6 e whenever 0 6 x - 1 6 d. 35–38. Challenging limit proofs Use the definition of a limit to prove the following results. 1 1 = (Hint: As x S 3, eventually the distance between x x 3 and 3 will be less than 1. Start by assuming x - 3 6 1 and 1 1 6 .2 show 2 x Copyright © 2014 Pearson Education, Inc. x - 16 = 8 (Hint: Factor and simplify.) xS4 x - 4 2 xSa a. For a given e 7 0, there is one value of d 7 0 for which f 1x2 - L 6 e whenever 0 6 x - a 6 d. b. The limit lim f 1x2 = L means that given an arbitrary d 7 0, 35. lim xS3 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 130 Limits x - 4 = 4 (Hint: Multiply the numerator and 1x - 2 denominator by 1x + 2.) 36. lim xS4 37. lim x S 1>10 1 = 10 (Hint: To find d, you will need to bound x away x Additional Exercises 44. The relationship between one-sided and two-sided limits Prove the following statements to establish the fact that lim f 1x2 = L if and only if lim- f 1x2 = L and lim+ f 1x2 = L. xSa xSa xSa xSa Assume f exists for all values of x near a with x 6 a. We say that the limit of f 1 x2 as x approaches a from the left of a is L and write lim- f 1x2 = L, if for any e 7 0 there exists d 7 0 such that xSa f 1x2 6 N whenever a 6 x 6 a + d. a. Write an analogous formal definition for lim+ f 1x2 = ∞ . xSa b. Write an analogous formal definition for lim- f 1x2 = - ∞ . xSa c. Write an analogous formal definition for lim- f 1x2 = ∞ . xSa 46–47. One-sided infinite limits Use the definitions given in Exercise 45 to prove the following infinite limits. 46. f 1x2 - L 6 e whenever 0 6 a - x 6 d. 39. Comparing definitions Why is the last inequality in the definition of lim f 1x2 = L, namely, 0 6 x - a 6 d, replaced xSa lim x S 1+ xSa 41. One-sided limit proofs Prove the following limits for 3x - 4 if x 6 0 f 1x2 = b 2x - 4 if x Ú 0. a. lim+ f 1x2 = - 4 xS0 xS0 c. lim f 1x2 = -4 xS0 42. Determining values of D from a graph The function f in the figure satisfies lim+ f 1x2 = 0 and lim- f 1x2 = 1. Determine a xS2 xS2 value of d 7 0 satisfying each statement. 6 6 6 6 2 1 2 1 whenever whenever whenever whenever 0 0 0 0 6 6 6 6 x x 2 2 - 2 2 x x 6 6 6 6 d d d d Use this definition to prove the following statements. 48. lim xS1 -2 = -∞ 1x - 122 49. lim x S -2 -10 = -∞ 1x + 224 50–51. Definition of a limit at infinity The limit at infinity lim f 1x2 = L means that for any e 7 0, there exists N 7 0 such that f 1x2 - L 6 e whenever x 7 N. Use this definition to prove the following statements. 50. lim 10 = 0 x 51. lim 2x + 1 = 2 x xS∞ xS∞ 52–53. Definition of infinite limits at infinity We say that lim f 1x2 = ∞ if for any positive number M, there is a xS∞ corresponding N 7 0 such that f 1x2 7 M whenever x 7 N. y Use this definition to prove the following statements. 6 52. lim x = ∞ 100 53. lim x2 + x = ∞ x xS∞ y ⫽ f (x) xS∞ 1 0 1 = ∞ 1 - x f 1x2 6 M whenever 0 6 x - a 6 d. xS∞ b. lim- f 1x2 = - 4 lim x S 1- xSa xSa 0 6 a - x 6 d in the definition of lim- f 1x2 = L? 47. any negative number M there exists a d 7 0 such that xSa 40. Comparing definitions Why is the last inequality in the definition of lim f 1x2 = L, namely, 0 6 x - a 6 d, replaced with 1 = -∞ 1 - x 48–49. Definition of an infinite limit We write lim f 1x2 = - ∞ if for with 0 6 x - a 6 d in the definition of lim+ f 1x2 = L? 0 0 1 1 xSa d 7 0 such that f 1x2 - L 6 e whenever 0 6 x - a 6 d. - xSa 45. Definition of one-sided infinite limits We say that lim+ f 1x2 = - ∞ if for any negative number N, there exists 39–43. Precise definitions for left- and right-sided limits Use the following definitions. Assume f exists for all x near a with x 7 a. We say that the limit of f 1 x2 as x approaches a from the right of a is L and write lim f 1x2 = L, if for any e 7 0 there exists d 7 0 such that f 1x2 f 1x2 f 1x2 f 1x2 xSa xSa 1 1 38. lim 2 = S x 5 x 25 a. b. c. d. xSa b. If lim f 1x2 = L, then lim- f 1x2 = L and lim+ f 1x2 = L. 1 1 ` 6 .) from 0. So let ` x 10 20 x S a+ xSa a. If lim- f 1x2 = L and lim+ f 1x2 = L, then lim f 1x2 = L. 1 2 6 43. One-sided limit proof Prove that lim+ 1x = 0. xS0 x 54. Proof of the Squeeze Theorem Assume the functions f, g, and h satisfy the inequality f 1x2 … g1x2 … h1x2 for all values of x near a, except possibly at a. Prove that if lim f 1x2 = lim h1x2 = L, then xSa xSa lim g1x2 = L. xSa Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Review Exercises 55. Limit proof Suppose f is defined for all values of x near a, except possibly at a. Assume for any integer N 7 0, there is another integer M 7 0 such that f 1x2 - L 6 1>N whenever x - a 6 1>M. Prove that lim f 1x2 = L using the precise xSa definition of a limit. 57. Prove that lim xS0 f 1x2 = e 56. For the following function, note that lim f 1x2 ≠ 3. Find a value xS2 of e 7 0 for which the preceding condition for nonexistence is satisfied. 0 if x is rational 1 if x is irrational. Prove that lim f 1x2 does not exist for any value of a. (Hint: xSa Assume lim f 1x2 = L for some values of a and L and let e = 12.) xSa f 1x2 - L 6 e whenever 0 6 x - a 6 d. x does not exist. x 58. Let 56–58. Proving that lim f 1x2 3 L Use the following definition xSa for the nonexistence of a limit. Assume f is defined for all values of x near a, except possibly at a. We say that lim f 1x2 ≠ L if for some e 7 0 there is no value of d 7 0 satisfying the condition xSa 59. A continuity proof Suppose f is continuous at a and assume f 1a2 7 0. Show that there is a positive number d 7 0 for which f 1x2 7 0 for all values of x in 1a - d, a + d2. (In other words, f is positive for all values of x in the domain sufficiently close to a.) QUICK CHECK ANSWERS 1 1. d = 50 or smaller 2. d = 0.62 or smaller decrease by a factor of 1100 = 10 (at least). y 131 3. d must ➤ 6 5 4 y ⫽ f (x) 3 2 1 0 CHAPTER 2 1. 1 2 3 4 x REVIEW EXERCISES Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. x - 1 has vertical asymptotes at a. The rational function 2 x - 1 x = -1 and x = 1. b. Numerical or graphical methods always produce good estimates of lim f 1x2. 2. Estimating limits graphically Use the graph of f in the figure to find the following values, if possible. a. f 1-12 b. lim - f 1x2 c. e. f 112 f. lim f 1x2 g. lim f 1x2 i. lim+ f 1x2 xS3 x S -1 xS1 y d. If lim f 1x2 = ∞ or lim f 1x2 = - ∞ , then lim f 1x2 does 6 xSa xSa not exist. e. If lim f 1x2 does not exist, then either lim f 1x2 = ∞ or xSa lim f 1x2 = - ∞ . x S -1 h. lim- f 1x2 xS2 xS3 xS3 c. The value of lim f 1x2, if it exists, is found by calculating f 1a2. xSa d. lim f 1x2 j. lim f 1x2 xSa xSa lim f 1x2 x S -1+ y ⫽ f (x) 5 xSa 4 xSa f. If a function is continuous on the intervals 1a, b2 and 1b, c2, where a 6 b 6 c, then the function is also continuous on 1a, c2. g. If lim f 1x2 can be calculated by direct substitution, then f is 3 2 xSa continuous at x = a. 1 ⫺1 Copyright © 2014 Pearson Education, Inc. 1 2 3 4 x

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