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```2.2 Limit of a Function Contemporary Calculus
2.2 THE LIMIT OF A FUNCTION
THE IDEA, Informally
Calculus has been called the study of continuous change, and the limit is the basic concept which allows us
to describe and analyze such change. An understanding of limits is necessary to understand derivatives,
integrals and other fundamental topics of calculus.
y mp
у—Кх) The limit of a function describes the behavior of the function when the variable 1s near,
U but does not equal , a specified number (Fig. 1). If the values of f(x) get closer
| and closer , as close as we want, to one number L as we take values of x very close
|
| A to (but not equal to) a number c, then we
с
Fig. 1
say "the limit of f(x), as x approaches c,is L " and we
write " lim f(x) =L." (The symbol " — " means "approaches" or "gets very close to.")
X—>C
f(c) is a single number that describes the behavior (value) of Г АТ the point x =c.
lim f(x) is a single number that describes the behavior of f NEAR, BUT NOT AT, the point x =c.
X—>C
If we have a graph of the function near x = c , then it is usually easy to determine lim f(x).
X—>C
Y y=1(x) ne
Example 1: Use the graph of y = f(x) in Fig. 2 to
determine the following limits:
(a) lim f(x) (b) lim f(x)
x—1 x—2
(с) lim f(x) (d) lim f(x)
x—3 x—4
Fig. 2
Solution: (a) lim f(x) =2 . When x is very close to 1, the values of f(x) are very close to
x—1
y = 2. In this example, it happens that {(1) = 2, but that is irrelevant for the limit. The only thing that
matters is what happens for x close to 1 but x #1.
(b) f(2) is undefined, but we only care about the behavior of f(x) for x close to 2 and not equal to 2. When
x 1s close to 2, the values of f(x) are close to 3. If we restrict x close enough to 2, the
values of y will be as close to 3 as we want, so lim f(x) = 3.
2.2 Limit of a Function Contemporary Calculus
(c) When x is close to 3 (or as x approaches the value 3), the values of f(x) are close to 1 (or
approach the value 1), so lim f(x) = 1. For this limit it is completely irrelevant that f(3) =2, We only
care about what happens to f(x) for x close to and not equal to 3.
(d) This one 1s harder and we need to be careful. When x is close to 4 and slightly less than 4 (x 1s just to
the left of 4 on the x-axis), then the values of f(x) are close to 2. But if x 1s close to 4 and slightly
larger than 4 then the values of f(x) are close to 3. If we only know that x is very close to 4, then we
cannot say whether y = f(x) will be close to 2 or close to 3 — it depends on whether x is on the right or
the left side of 4. In this situation, the f(x) values are not close to a single number so we say
lim f(x) does not exist. It is irrelevant that f(4) = 1. The limit, as x approaches 4, would still
x—4
be undefined if f(4) was 3 or 2 or anything else.
y f(x)
Practice 1: Use the graph of y = f(x) in Fig. 3 to determine
the following limits:
(a) lim f(x) (b) lim f(t)
I (с) lim f(x) (d) lim f(w)
Fig. 3 x—3 w—4
2x —x-—1
Example 2: Determine the value of lim — 7
x—3 X —
Solution: We need to investigate the values of f(x) when x 1s close to 3. If the f(x) values get
arbitrarily close to or even equal some number L, then L will be the limit. One way to keep track of both
the x and the f(x) values is to set up a table and to pick several x values which are closer and closer (but
not equal) to 3. We can pick some values of x which approach 3 from the left, say x =2.91,2.9997,
2.999993, and 2.9999999 , and some values of x which approach 3 from the right, say x = 3.1, 3.004,
3.0001, and 3.000002 . The only thing important about these particular values for x 1s that they get closer
and closer to 3 without equaling 3. You should try some other values "close to 3" to see what happens.
Our table of values is
X f(x) X f(x)
29 6.82 3.1 72
2.9997 6.9994 3.004 7.008
2.999993 6.999986 3.0001 7.0002
2.9999999 6.9999998 3.000002 7.000004
y y | y
3 7 3 7
As the x values get closer and closer to 3, the f(x) values are getting closer and closer to 7. In
fact, we can get f(x) as close to 7 as we want ("arbitrarily close") by taking the values of x very
| ‚ 2х’ -х-1
close ("sufficiently close") to 3. lim 1 =
x—3 xX —
7.
Instead of using a table of values , we could have graphed y = f(x) for x close to 3 , Fig. 4, and used
2
2.2 Limit of a Function
Contemporary Calculus
the graph to answer the limit question. This graphic approach is easier, particularly if you have a
calculator or computer do the graphing work for you, but it is really very similar to the "table of values"
method: in each case you need to evaluate y = f(x) at many values of x near 3.
You might have noticed that if we just evaluate f(3), then we get the correct answer 7. That works for this
particular problem, but it often fails. The next example illustrates the difficulty.
Fig. 4
1
xXx ==]
x-]
---- Near x=3
note the
solid dot
X
. . x —x—1
Example 3: Find lim . (Same as Example 2 but
x—1 Х —
with x—1))
2x2_x_1
Solution: You might try to evaluate f(x)= Ор] — at x=1,
but f is not defined at x = 1. It is tempting, but wrong, to
conclude that this function does not have a limit as x approaches 1.
Table Method: Trying some "test" values for x which get closer and closer to 1 from both the left
and the right, we get
The function f is not defined at x = 1, but when x 1s close to 1, the values of f(x) are getting very close to 3.
We can get f(x) as close to 3 as we want by taking x very close to 1 so lim
X f(x) X f(x)
09 2.82 1.1 3.2
0.9998 2.9996 1.003 3.006
0.999994 2.999988 1.0001 3.0002
0.9999999 2.9999998 1.000007 3.000014
| | | |
1 3 1 3
2х” -х -1
————— =3
x—1 x-1
2x2 _ x —
1
Graph Method: We can graph y=1(x)= —_-7 for x closeto 1, Fig. 5, and notice that
whenever x 1s close to 1, the values of y = f(x) are close to 3. f is not defined at x = 1, so the graph has
a hole above x = 1, but we only care about what f(x) is doing for x close to but not equal to 1.
Fig. 5
GE
¥ —
х- |
2
2х -х-1 Algebra Method: We could have found the same result by noting
2x2 — x —1 (2х+1)(х-1)
E =
that f(x) = х — 1 = —
= 2x+1 as long as
——__ hear x=]
note the
open dot
X
x #1. (If x#1, then x-1 #0 so it is valid to divide the
numerator and denominator by the factor x—1.) The "x—>1"
part of the limit means that x is close to 1 but not equal to 1,
so our division step 1s valid and
2х” -х -1
Шо — — = lim 2x +1 =3, the correct answer.
x— Х — x—
WSBCTC
3
2.2 Limit of a Function Contemporary Calculus
THREE METHODS FOR EVALUATING LIMITS
The Algebra Method
The algebra method involves algebraically simplifying the function before trying to evaluate its limit.
Often, this simplification just means factoring and dividing, but sometimes more complicated algebraic or even
trigonometric steps are needed.
The Table and Graph Methods
To evaluate a limit of a function f(x) as x approaches c, the table method involves calculating the values of
f(x) for "enough" values of x very close to c so that we can "confidently" determine which value f(x) 1s
approaching. If f(x) 1s well-behaved, we may not need to use very many values for x. However, this method 1s
usually used with complicated functions, and then we need to evaluate f(x) for lots of values of x.
A computer or calculator can often make the function evaluations easier, but their calculations are subject
to "round off" errors. The result of any computer calculation which involves both large and small numbers
should be viewed with some suspicion. For example, the function
019 +1y-1 (03
(0.1) (0.1)
f(x) = = 1 for every value of x , and my calculator gives the correct
£(0.07+13-1
(0.1?
answer for some values of x: f(3) = = 1,and f(8) and f(9) both equal |.
But my calculator says { (0.1) +1 }—1 =0 so it evaluates f(10) to be 0, definitely an incorrect value.
Your calculator may evaluate f(10) correctly, but try f(35) or f(107).
Calculators are too handy to be ignored, but they are too prone
to these types of errors to be believed uncritically. Be careful.
The graph method 1s closely related to the table method, but we create a graph of the function instead of a table
of values, and then we use the graph to determine which value f(x) 1s approaching.
Which Method Should You Use?
In general, the algebraic method 1s preferred because it 1s precise and does not depend on which values of x we
chose or the accuracy of our graph or precision of our calculator. If you can evaluate a limit algebraically,
you should do so. Sometimes, however, it will be very difficult to evaluate a limit algebraically, and the table
or graph methods offer worthwhile alternatives. Even when you can algebraically evaluate the limit of a
function, it 1s still a good idea to graph the function or evaluate it at a few points just to verify your algebraic
The table and graph methods have the same advantages and disadvantages. Both can be used on very
complicated functions which are difficult to handle algebraically or whose algebraic properties you don't know.
2.2 Limit of a Function Contemporary Calculus
Often both methods can be easily programmed on a calculator or computer. However, these two methods are
very time—consuming by hand and are prone to round off errors on computers. You need to know how to use
these methods when you can't figure out how to use the algebraic method, but you need to use these two methods
warily.
Xx +5x+6 Xx +5x+6
Example 4: Evaluate (a) lim ——— and (by Im —/——.
x0 x2 + 3x42 x>2 XxX +3X +2
Solution: The function in each limit 1s the same but x 1s approaching a different number in each of them.
(a) Since x—0, we know that x is getting closer and closer to Ô so the values of the x2 ‚5x and 3x terms get
as close to 0 as we want. The numerator approaches 6 and the denominator approaches 2, so the values
of the whole function get arbitrarıly close to 6/2 = 3, the limit.
(b) Asx approaches —2, the numerator and denominator approach 0, and a small number divided by a
small number can be almost anything — the ratio depends on the size of the top compared to the bottom.
More investigation is needed.
Table Method: If we pick some values of x close to (but not equal to) —2, we get the table
2 2 x2 + 5x +6
X X“+5x +6 X“+3x +2 D a a
x“ +3x +2
—1.97 0.0309 —0.0291 —1.061856
—2.005 —0.004975 0.005025 —0.990050
—1.9998 0.00020004 —0.00019996 —1.00040008
—2.00003 —0.00002999 0.0000300009 —0.9996666
| | | |
—2 0 0 —1
Even though the numerator and denominator are each getting closer and closer to 0, their ratio is
getting arbitrarily close to —1 which is the limit.
Graph Method: The graph of y = f ARTO pes
: = = "О .
я 2 | y ra etho e graph of y (x) 2+ 3x +2 in Fig
| H shows that the values of f(x) are very close to —1 when the
|
Bo Be | x—values are close to —2.
a a E „1 E
хо +3х+2 | x2 ZeSx+6 _ (42)
(к+2)/(х + 3) Algebra Method: f(x) = 243x412 = FFD) -
TALE. A оо =
(x+2)(x +1) | We know x — —2 so x # —2, and we can divide the top
Fig. 6 and bottom by (x+2). Then f(x) = (x+3)/(x+1) so
f(x) > 1/+1 =—1 as x > 2.
polynomial
0
If lim {polynomial polynomial } approaches 75 , try dividing the top and bottom by x-c.
(cc) (+) | WSBCTC
2.2 Limit of a Function Contemporary Calculus
O X-X—2 . t-sin(f) , W-2
Practice2: Evaluate (a) lim ——— (b) lim ——— (с) lim——.
x>2 x-2 1>0 1° +3t w=2 In(w/2)
ONE-SIDED LIMITS TA
т e——o
Sometimes, what happens to us at a place depends on the direction we use to ara
approach that place. If we approach Niagara Falls from the upstream side, then we
will be 182 feet higher and have different worries than if we approach from the .
downstream side. Similarly, the values of a function near a point may depend on the
direction we use to approach that point. If we let x approach 3 from the left (x 1s Fig. 7
close to 3 and x < 3), then the values of [x] =INT(x) equal 2 (Fig.7). If we let x
approach 3 from the right (x 1s close to 3 and x > 3), then the values of [x] =INT(x) equal 3.
On the number line we can approach a point from the left or right, and that leads to one-sided limits.
Definition of Left and Right Limits:
The left limit as x approaches C of f(x) is L 1f the values of f(x) get as close to Las we
want when x is very close to and left of c, x <c: lim f(x) =L.
X—>C
+
The right limit, written with x =» e ,requires that x lie to the right of €, x > c.
Example 5: Evaluate lim (x — [x] ) and lim (x — [x] ).
x>2"
x>2
Solution: The left-limit notation x — 2 requires that x be close to 2 and that
x be to the left of 2, so x <2.
If 1<x<2,then [x]=1 so lim (x-[x])=2-1=1.
x>2
X
If x 1s close to 2 and is to the right of 2,then 2 <x <3 so [x]=2
and lim (x | х ) = 2-2 = 0.
x—>2*
The graph of f(x) = x — [x] 1s shown in Fig. 8 .
If the left and right limits have the same value, lim f(x) = lim f(x) = L, then the value of f(x) is close to
L whenever x is close to c, and it does not matter if x is left or right of ¢ so lim f(x) =L. Similarly, if
X—>C
lim f(x) = L , then f(x) is close to L whenever x is close to c and less than c and whenever x is close to ¢
X—>C
and greater than c, so lim f(x) = lim f(x) =L . We can combine these two statements into a single
x>c x>c"
theorem.
2.2 Limit of a Function Contemporary Calculus
One-Sided Limit Theorem:
lim f(x)=L ifandonlyif lim f(x) = lim f(x)=L
X—>C
Corollary: If lim f(x) = lim f(x), then lim f(x) does not exist.
One-sided limits are particularly useful for describing the behavior of functions which have steps or jumps.
To determine the limit of a function involving the greatest integer or absolute
value or a multiline definition, definitely consider both the left and right limits.
Practice 3: Use the graph in Fig. 9 to evaluate the one and two-sided
limits of f at x=0,1,2, and 3.
1 if x <1
* Practiced: Let f(x) = > A
if 3 <x
Find the one and two-sided limits of f at 1 and 3.
PROBLEMS y
EE A-|
1. Useth h in Fig. 10 to det ine the following limits. | | | |
se the graph in Fig o determine the following limits A è
@ lim f(x) © lm f(x) jUN 44,
Lt 2 NU 4
. . -1+ = ван ANA ps]
© lim f(x) a) lim f(x) то
у Кх) on os
2. Use the graph in Fig. 11 to determine the following limits.
(2) lim f(x) (b) lim f(x)
© lim f(x) (@) lim f(x)
2.2 Limit of a Function
3. Use the graph in Fig. 12 to determine the following limits.
(a)
lim 2х) (b)
lim f2x-5) — (d)
5. Evaluate
6. Evaluate
7. Evaluate
8. Evaluate
9. Evaluate
10. Evaluate
11. Evaluate
X
12. f(x) = | sin)
ON 00 —
13. g(x) = |
Et
/X if 2<x<4
x>1 x-2
(a) lim x+7
a —o—————————————;.—O<—
x>0 Хх” + 9х + 14
x>4 x*+9x+14
cos(x)
(a) lim
x—1 X
(a) lim \х - 3
(a) lim |x|
x—0"
xl
(a) lim —
x>0 X
(a) lim IXx—3|
if x<0
if O<x<2
if 2<x
if x <2
—X if 4 <x
lim f(x-1)
x—2
lim f(4+x)
x—0
Contemporary Calculus
the following limits.
(a) lim f(3x) (b)
x—
(c) lim f(2x—4) (d)
4. Use the graph in Fig. 13 to determine
lim f(x+1)
x—2
lim | f(4+x) |
x—0
6) lim XxX +3x+3
x>2 Xx—2
(b) lim xT
x>3 x“ +9x +14
(d) lim e
x>7 Xx +9x +14
X—>J X x—-1 X
() lim Vx =3
lim
(с) 1-9
Vx -3
(b) lim |x| (с) lim | x |
x>0" x—
xl 2 1х!
(b) lim — (c) Im —
x—0" X x>0 x
2 |х —5| 2 |х-5!
(b) lim (c) lim
x=3 x-5 x>5 x-5
. Find the one and two-sided limits of f as x = 0,1, and 2.
. Find the one and two-sided limits of g as x > 1,2,4,and 5.
WSBCTC
8
2.2 Limit of a Function Contemporary Calculus 9
In problems 14 — 17 use a calculator or computer to get approximate answers accurate to 2 decimal places.
2-1 . log (x , — .In(x
14. (a) lim (b) lim log, (x) 15. (a) lim (b) lim (x)
x—0 X x—1 XxX — x—0 X x>» x — 1
o Vx-1-2 — sin(3x | Мх -4 ‚| sin(7x
16. (а) lim — —— (b) lim Sin(3x) 17. (а) lim — —— (b) lim sin(7x)
x>5 X-5 x>0 5x x>16 Xx—16 x=0 2x
? |
18. Define A(x) to be the area bounded by the x and y axes, the ЧТ / v=f(1)
bent line in Fig. 14, and the vertical line at x. For example, it
A(4) = 10. su <— A(x)
= area
a) Evaluate А(0), А(1), А(2), апа А(3). =
b) Graph y=A(x) for 0<x=<4. ; | |
| | I | |
с) What area does A(3) — A(1) represent” 1 2 3 - 4 3
Fig. 14
19. Define A(x) to be the area bounded by the x and y axes, the 1
1 y y = —t+2
line y = 5 x +2 , and the vertical line at x. (Fig. 15). 4-1 E =
|
For example, A(4) = 12. IT
ая <— AG)
a) Evaluate А(0), А(1), А(2), апа А(3). as
b) Graph y = A(x) for 0 <x <4, ı +
с) What area does A(3) — A(1) represent” | | | | |
1 2 13 4 5
Fig. 15
20. Sketch f(x)=V4x-x" for 0 <x <4 (this is a semicircle) .
Define A(x) to be the area bounded by the x and y axes, the graph y = f(x) , and the vertical line at x.
a) Evaluate A(0), A(2),and A(4).
b) Graph y = A(x) for 0 <x <4,
с) What area does A(3) — A(1) represent”
GE
WSBCTC
9
2.2 Limit of a Function
Practice 1: (a) 2
Practice 2: (a)
(b)
(с)
Practice 3: lim
x—0"
lim
x>l
lim
x>2
lim
x>3
Practice 4: lim
x>l
lim
x>3
GE
lim
x—>2
lim
t—0
w—2
f(x)
f(x)
f(x)
f(x)
f(x)
f(x)
Contemporary Calculus
10
(b) 2 (c) does not exist (no limit) (d) 1
x+D(x-2
( X ) = lim (x+1) = 3
х -2 x>2
t - sin(t sin(7 0
msm _ im (1) = — = 0
t(t+ 3) 170 (+3 3
W-2 |
2. Try this one numerically or using a graph.
In(w/2)
\ — 2 \ — 2
w In( w/2 ) w In( w/2 )
2.2 2098411737 1.9 1.949572575
2.01 2004995844 1.99 1.994995823
2.003 2001499625 1.9992 1999599973
2.0001 2.00005 1.9999 1.99995
1 lim f(x) = 2 lim f(x) does not exist
x—.0* x—
1 lim f(x) = 1 lim f(x) 1
x—1" х->1
—1 lim f(x) = -1 lim f(x) = -1
x—>2" x—2
—1 lim f(x) = 1 lim f(x) does not exist
x—3" x—
1 lim f(x) = 1 lim f(x) =1
x—1" х->1
3 lim f(x) = 2 lim f(x) does not exist
x—3" x—
WSBCTC 10
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