MAT137Y1a.doc Lecture #1 – Tuesday, September 9, 2003 SETS • • non-example: A = {paintings that are beautiful} example: A = {natural numbers divisible by 4} Notation • • 4 ∈ A – “4 is in A” 3 ∉ A – “4 is not in A” • B = {natural numbers divisible by 7} A ∪ B = {natural numbers divisible by 4 or 7} • – “union of A and B” = {4,7,8,12,14,...} • A ∩ B = {natural numbers divisible by 4 and 7} – “intersection of A and B” FACTS ABOUT REAL NUMBERS • • • • N = natural numbers = {1,2,3,4,...} Z = integers = {...,−3,−2,−1,0,1,2,3,...} Q = rational numbers = {1,2,3,4,...} R = real numbers = all this and more (+ irrational numbers) Geometrically -5 • -4 -3 -2 -1 0 1 2 3 4 any point on the line represents a real number Intervals • • • [− 1,2] = {x ∈ R : −1 ≤ x ≤ 2} (− 1,2) = {x ∈ R : −1 < x < 2} [0, ∞ ) Ordering (Inequalities) • If a, b are real numbers, then exactly one of the following is true: Important Properties • • If a < b and c > 0 , then ac < bc If a < b and c < 0 , then ac > bc Page 1 of 33 a<b a=b a>b 5 The Real Line MAT137Y1a.doc Distance/Absolute Values a, if a ≥ 0 • a = distance from a to 0 = • a − b = distance between a and b − a, if a < 0 Triangle Inequality • • a+b ≤ a + b analogous to triangle theorem from geometry a b c c ≤ a+b c • Lecture #2 – Thursday, September 11, 2003 TRIANGLE INEQUALITY • a + b ≤ a + b , a, b ∈ R Proof (A Proof By Cases) 1) If a, b both ≥ 0 • Then a + b ≥ 0 a+b = a+b = a + b • 2) If a > 0, b < 0, a ≥ b • • 3) If 4) Then a + b ≥ 0 a+b = a+b a + b = a −b • Because b < 0 , so a + b < a − b • • Then a + b ≥ 0 a+b = a+b = a + b a > 0, b < 0, a < b 5) 6) Page 2 of 33 MAT137Y1a.doc REVIEW OF INEQUALITIES Solve 1) 2x + 3 ≤ 6 2x ≤ 3 3 x≤ 2 −2 x + 3 ≤ 5 − 2x ≤ 2 x ≥ −1 x ∈ [− 1, ∞ ) 2) 3 x ∈ − ∞, 2 (x − 1)(x − 4)(x + 2) > 0 Where is it = 0? x = 1,4,−2 3) (x − 1) (x − 4 ) (x + 2 ) -2 x <δ • x−c <δ - x 0 c- c - + + - - - + - + + + Product + ∴ x ∈ (− 2,1) ∪ (4, ∞ ) - + −δ < 0 < δ x ∈ (− δ , δ ) – it means that x c+ – it means that x −3 > 5 • x−3 > 5 -2 3 8 x − 3 < −5 ∴ (− ∞,−2) ∪ (8, ∞ ) COORDINATE GEOMETRY Rectangular Coordinates y-axis 3 2 1 4 - Inequalities With Absolute Values • 1 (3,1) 1 2 3 x-axis Page 3 of 33 c −δ < x < c +δ x ∈ (c − δ , c + δ ) MAT137Y1a.doc Pythagorean Theorem (x1, y1) |y1 – y0| (x0, y0) |x1 – x0| d= x1 − x 0 2 + y1 − y 0 2 Line Equations • y − y 0 = m(x − x 0 ) – “point-slope” form y = mx + b – “y-intercept” form ax + by + c = 0 – standard form • If • • y = m1 x + b1 y = m 2 x + b2 , • Two lines are parallel if m1 = m 2 • Two lines are perpendicular if m1 m 2 = −1 Conic Sections • • ax 2 + by 2 + cx + dy + e = 0, a, b, c, d , e are constants Get ellipses, hyperbolas, parabolas • (x − 2)2 + ( y − 3)2 = 9 ( x − 2 ) 2 + ( y − 3) 2 = 3 3 (2,3) 3 3 FUNCTIONS • A rule (“black box”) that takes an input and produces a single output Page 4 of 33 MAT137Y1a.doc Examples • f ( x) = x 2 • Domain of f is the set of legitimate inputs – domain ( f ) = (− ∞, ∞ ) • Range of f is the set of legitimate outputs – range( f ) = [0, ∞ ) • f ( x) = 9 − x 2 • domain ( f ) = [− 3,3] range( f ) = [0,3] • Piecewise Defined Functions • x 2 if x ≥ 0 − x if x < 0 f (x ) = GRAPHS AND FUNCTIONS • graph ( f ) = {( x, y ) : x ∈ dom( f ), y = f (x )} Example f (x ) = x 2 + 1 domain ( f ) = (− ∞, ∞ ) range( f ) = [1, ∞ ) (0,1) SOME BASIC FUNCTIONS Polynomials • f (x ) = a n x n + a n −1 x n −1 + ... + a1 x + a 0 • ai are constants • domain ( f ) = (− ∞, ∞ ) Rational Functions • f (x ) = • P(x ) , where P, Q are polynomials Q( x ) domain ( f ) = {x ∈ R : Q( x ) ≠ 0} Page 5 of 33 MAT137Y1a.doc TRIGONOMETRIC FUNCTIONS (x,y) 1 θ (1,0) sin θ = y cos θ = x y tan θ = x 1 csc θ = y 1 sec θ = x x cot θ = y Example of a Trigonometric Identity • sin 2 θ + cos 2 θ = 1 , because x 2 + y 2 = 1 COMBINING FUNCTIONS • If f, g are functions • “sum” – ( f + g )(x ) = f (x ) + g ( x ) • • • ( f − g )(x ) = f (x ) − g (x ) “product” – ( f ⋅ g )(x ) = f (x ) ⋅ g (x ) Warning: domain is domain ( f ) ∩ domain (g ) “difference” – Example • Given f (x ) = 1 + x and f (x ) = 1 + x , g ( x ) = − x + 5 domain ( f ) = [− 1, ∞ ) • domain ( g ) = (− ∞,5] • domain ( f + g ) = [− 1,5] COMPOSITION OF FUNCTIONS (TWO “BLACK BOXES” IN SUCCESSION) • • (f g )(x ) = f (g ( x )) Warning: f g ≠ g f Examples • Given f (x ) = x + 3 and g ( x ) = x 2 • • • • (f (g g )(x ) = x 2 + 3 f )(x ) = (x + 3)2 domain ( f g ) = {x ∈ domain ( g ) : g (x ) ∈ domain ( f )} Given f (x ) = x − 2 and g (x ) = x • domain ( f g ) = [4, ∞ ) Page 6 of 33 MAT137Y1a.doc Lecture #3 – Tuesday, September 16, 2003 INTRODUCTION TO PROOFS: “PROOFS ARE YOUR FRIENDS” Mathematical Rigor: To What End? • Moral: Need rigor to obtain a correct, elegant solution which doesn’t come from brute force Proofs: To What End? • Informally, a proof is a guarantee that a claim is true Definitions • • • Proposition – a sentence that is either true or false Theorem – a proposition that is guaranteed by a proof Proof – showing a theorem follows logically from the set of axioms – ex: B follows logically from A if B is true in all possible worlds where A is true Examples 1) The acceleration of a rigid body is proportional to force applied. • False is real world 2) For all integers n, if n > 2 , there are no positive integers a, b, c such that a n + b n = c n • True 3) For all non-negative integers n, n 2 + n + 41 is prime • False – n = 41 would make it false (counterexample) Notation • • ∀ – “for all” – universal quantifier ∃ – “there exist”, “for some” – existential quantifier P Q False False True True False True False True ¬Q not Q True True False False P∧Q P and Q False False False True P∨Q P or Q False True True True I: PROOF BY ENUMERATION Example 1 • • • Given: Roses are red and violets are blue. Prove: Roses are red. Proof: 4th line from the table ( P ∧ Q ) Example 2 • • Given: If it rains today, I’ll eat my hat. It rains today. Prove: I’ll eat my hat. Page 7 of 33 P Q if P, then Q True True False True P⇔Q P iff Q True False False True ¬Q ¬P True True False True MAT137Y1a.doc • Proof: 4th line from the table ( P Q) Lecture #4 – Thursday, September 18, 2003 II: PROOF BY CONTRAPOSITIVE (“INDIRECT PROOF”) Example 1 • If John is at work, then he’s logged in. ( P • If John is not logged in, then he’s not at work. ( ¬Q • • Warning: Converse of P Q is Q P Moral: Proving contrapositive is logically equivalent to proving the original statement Q) ¬P ) Example 2 Prove: For any integer n, if n2 is even, then n is even. Proof: • Reformulate: If n is odd, then n2 is odd. 1) If n is odd, then n = 2a + 1, a ∈ Z ( ) 2) If n = 2a + 1 , then n 2 = (2a + 1)2 = 4a 2 + 4a + 1 = 2 2a 2 + 2a + 1 3) Because a ∈ Z , 2a 2 + 2a ∈ Z 4) Because 2a 2 + 2a ∈ Z , n2 is odd QED Example of A Non-Proof Prove: The “Theorem” 1 = −1 Proof: (− 1)(− 1) = −1 ⋅ −1 = ( −1) • 1= 1 = • • But a ⋅ b ≠ a ⋅ b for all a and b! Moral: Justify each step. 2 = −1 QED Other Classical Errors • • Divide both sides of an equation by a variable – ax = bx a = b – what if x = 0 ? Divide both sides of an inequality by a variable – ax < bx a < b – what if x ≤ 0 ? III: PROOF BY CONTRADICTION • • Want to show (WTS): P Idea: Assume P is not true, then show that a contradiction occurs when combining it with the axioms. Example Prove: 2 is irrational. Page 8 of 33 MAT137Y1a.doc Proof: 1) Assume that 2 is rational. That means ∃a, b ∈ Z such that 2= a , where a, b have no common b factors. a2 , so 2b 2 = a 2 b2 3) That means a2 is even. So by theorem above, a is even. 4) a = 2k for some k ∈ Z , so a 2 = 4k 2 . 2) Then 2 = 5) So 2b 2 = 4k 2 , so b 2 = 2k 2 . 6) So b2 is even by theorem above. 7) 2 divides both a and b. 8) Contradiction: a, b have no common factors. Thus, • 2 is irrational. lemma(s) – smaller theorems used to prove bigger theorems IV: PROVE BY CASES • Warning: It only works if there are a finite number of cases. Example Prove: The Triangle Inequality a + b ≤ a + b , a, b ∈ R • • • Proof: Case 1: a, b ≥ 0 … Case 2: a, b < 0 … Case 3: a ≥ 0, b < 0, a ≥ b … • Case 4: a ≥ 0, b < 0, a < b … V: PROOF BY INDUCTION • • What do you do when there are infinitely many cases? Example: Universally quantified statements over N – i.e. “for all n ∈ N …” Principle of Induction • If you can prove that: some property holds for n = 0 (“base case”) and if the property holds for n = k , it holds for n = k + 1 (“inductive step”), then the property holds for all n ∈ N . Example Prove: ∀n ∈ N , n 3 − n is divisible by 3. Definition: For all a, b ∈ Z , “a divides b” ( a | b ) iff ∃q ∈ Z, b = a ⋅ q Proof: • “Base Case”: If n = 0 , the 0 3 − 0 = 0 . 3 | 0 because 0 = 3 ⋅ 0 • “Inductive Step”: Suppose it is true for n = k so 3 | k 3 − k and k 3 − k = 3q for some q ∈ Z . Page 9 of 33 MAT137Y1a.doc • Now WTS: 3 | (k + 1)3 − (k + 1) (k + 1)3 − (k + 1) = k 3 + 3k 2 + 3k + 1 − k − 1 = k 3 − k + 3k 2 + 3k ( = 3q + 3 k 2 + k ) So, by definition of divisibility, 3 | (k + 1)3 − (k + 1) QED Non-Proof Example “Theorem”: All iMacs are the same colour. Prove: All n ∈ N , a set of iMacs of size n is monochromatic. • • Proof: “Base Case”: n = 1 , then every set of one iMacs is one colour “Inductive Step”: Suppose true for n = k , WTS for n = k + 1 Let S = {i1 , i 2 ,...i k +1 } S1 = {i1 , i 2 ,...i k } , S 2 = {i 2 , i 3 ,...i k +1 } S1 ∩ S 2 = {i 2 ...i k } • Since S1, S2 are monochromatic and they intersect in S1 ∩ S 2 , so S is monochromatic. So all iMacs are monochromatic. QED • But the “Inductive Step” fails for n = 2 VI: STRONG INDUCTION • Same as Induction, except n = k is replaced with “if the property holds for n ≤ k …” Example (Half of “Fundamental Theorem of Arithmetic”) Prove: Any natural number n, for n > 1 , can be written as a product of primes Proof (Attempt with Induction): • “Base Case”: n = 2 . n is a prime. • “Inductive Step”: Suppose that n = k , k can be written as a product of primes, WTS for n = k + 1 STUCK – primes that divide k don’t divide k + 1 • • Proof (Attempt with Strong Induction): “Base Case”: n = 2 . n is a prime. “Inductive Step”: Suppose that n ≤ k , k can be written as a product of primes, WTS for n = k + 1 • Case 1: n = k + 1 is prime. • Case 2: n = k + 1 is not prime, so ∃a, b ∈ N,1 < a, b < k + 1 so that k + 1 = a ⋅ b . Because a, b ≤ k , so a ⋅ b can be written as a product of primes. Therefore, k + 1 is a product of primes • Therefore, by strong induction, it is true for all n ∈ N . QED Page 10 of 33 MAT137Y1a.doc Lecture #5 – Tuesday, September 23, 2003 WHY CALCULUS? • Bascially, because the study of natural phenomena involves: • rates of change • non-linear phenomena – calculus allows for linear approximation LIMITS Basic Idea 1) A (precise) way to take successive approximations and get an exact answer 2) Start with things you can easily compute (ex: lengths of line segments, area or rectangles) and build up to computations for more complicated objects Pictures 1) Length of a (non-linear) curve: 2) Volume of a solid: The Question What happens to a function value f (x ) as x approaches a fixed number c? Example • • f (x ) = x − 1 c=2 The function value then approaches 1 – lim(x − 1) = 1 x →1 f(x) f(x) = x - 1 y=2 x c=2 Notation lim f (x ) = L x→c Page 11 of 33 MAT137Y1a.doc Definition (Attempt 1) • f (x ) gets closer to L as x gets close c Example • f (x ) = x 2 f(x) c = −2 f(x) = x2 L=4 x c = -2 Technical Note Doesn’t matter whether the function is defined at x = c or not – ask only about values of x close to c, but not equal c Example • • x 2 −1 x +1 c = −1 f(x) f (x ) = f(x) = x2 – 1 x+1 x 2 −1 = −2 x → −1 x + 1 c = -1 lim x L = -2 Example Why is something not the limit? f (x ) = x • c =1 f(x) f(x) = x L=2 L=2 x c=1 • What went wrong? f ( x ) − 2 > 0.5 if x − 1 < 0.5 Definition (Attempt 2) Page 12 of 33 x f (x ) − 2 1.5 1.25 1.1 0.499 0.249 0.099 MAT137Y1a.doc • lim f (x ) = L if no matter what distance ε (on y-axis) I pick, if I say x→c f ( x ) − L < ε , then you should be able to find another distance δ (on x-axis) so that you can guarantee that if x − c < δ , then y f (x ) − L < ε f(x) L+ε y=L L-ε x c-δ x=c c+δ Lecture #6 – Thursday, September 25, 2003 LIMITS (CONTINUED) Definition (Attempt 3 and Final): Epsilon-Delta Definition of Limits • • Let f (x ) be defined in some interval of the form (c − p, c ) ∪ (c, c + p ) for some p > 0 We say lim f (x ) = L if for all ε > 0 , there exist a δ > 0 such that if x − c < δ then x→c ∀ε > 0, ∃δ > 0 such that x − c < δ f (x ) − L < ε – f (x ) − L < ε Example f (x ) = 2 x + 1 If ε is equal to… ε =1 y c =1 L=3 (2 x + 1) − 3 < 1 ε = 10 Any ε y=1 2x − 2 < 1 2 x −1 < 1 x −1 < 1 2 1 c=1 Can choose δ to be… 1 δ= 2 1 δ= 20 ε δ= 2 x Technical Note • • • Warning: The limit of the function at x = c can be different from the function value f (c ) A limit exists exactly when both left hand side and right hand side limits exist and they are equal 1 Another way that limits might not exist – ex: f (x ) = , c = 0 x Page 13 of 33 MAT137Y1a.doc Definition Left hand limit: • Assume f is defined on some interval (c − p, c ) for p > 0 Then lim− f (x ) = L if ∀ε > 0, ∃δ > 0 such that c − δ < x < c • x →c Right hand limit: • Assume f is defined on some interval (c, c + p ) for p > 0 Then lim+ f (x ) = L if ∀ε > 0, ∃δ > 0 such that c < x < c + δ • x →c f (x ) − L < ε f (x ) − L < ε Example Prove (using ε-δ) that lim x 2 = 4 x→ 2 • Let ε > 0 • Want: x 2 − 4 < ε – our task is to find a δ so that x − 2 < δ then x 2 − 4 < ε holds • Algebra: x 2 − 4 = (x + 2 )(x − 2 ) • Trick: From now on, assume we’ll take δ < 1 That means we can assume x − 2 < 1 so 1 < x < 3 • That means • Choose δ = min • Check: If δ < ( ) x+2 < x +2 x+2 <5 , so x 2 − 4 = x + 2 x − 2 < 5 x − 2 ε ,1 5 ε ε , then x 2 − 4 < 5 x − 2 < 5 <ε 5 5 Q.E.D. Example Prove: lim x = 2 x→ 4 • Want: For ε > 0 , • By the properties of the square root function, we already know δ < 4 – so 0 < x < 8 • Algebra: x−4 = • 2< x − 2 < ε for small enough x − 4 (x − 4 ) = ( x +2 x +2 )( x −2 ) x −2 x + 2 because x > 0 Page 14 of 33 MAT137Y1a.doc 1 x−4 2 x −2 < • • Choose: δ = min (4,2ε ) • Check: If x − 4 < δ , then x −2 < 1 1 x − 4 < (2ε ) = ε 2 2 Q.E.D Example 1 Prove: lim x→ 2 x 2 1 4 = 1 Want: For ε > 0 , want • We already know δ < 2 , 0 < x < 4 • Algebra: • • • 1 x 2 − 1 x2 − x 2 − 1 < ε for x close enough to 2 (ie: x − 2 ) 4 • 1 4 − x 2 (2 + x )(2 − x ) = = 4 4x 2 4x 2 1 (2 + x )(2 − x ) = (2 + x )(2 − x ) = (2 + x ) x − 2 = 4 4x 2 4x 2 4x 2 Make additional requirement: δ < 1 , so 1 < x < 3 2+ x < 2+ x < 2+3< 5 1 4x 2 < 1 because x > 1 4 (2 + x ) < 5 4 • Therefore, • So • Choose δ = min 1, • Check: If x − 2 < δ , then 1 x2 − 4x 2 1 5 < x−2 4 4 4 ε 5 1 x 2 − 1 5 < x−2 < ε 4 4 Page 15 of 33 MAT137Y1a.doc Lecture #7 – Tuesday, September 30, 2003 COMPUTATIONS WITH LIMITS Building Blocks 1) 2) 3) lim x = c x →c lim x → 0 = x c lim k = k x→c Theorem 1 • If lim f (x ) = L and lim g (x ) = M x→c x →c • lim ( f ( x ) + g (x )) = L + M • lim (αf ( x )) = α ⋅ L • lim ( f (x ) ⋅ g (x )) = L ⋅ M x →c x→c x→c Proof 1 Given ε > 0 , we want to find δ > 0 so that if x − c < δ then • Algebra: f ( x ) + g (x ) − L − M = ( f ( x ) − L ) + (g (x ) − M ) ≤ f ( x ) − L + g (x ) − M • We know that lim f (x ) = L and lim g (x ) = M • If • f ( x ) + g ( x ) − (L + M ) < ε x→c x →c ε ε and g (x ) − M < , then we’re in business! 2 2 ε ε ε For > 0 , we can find δ1 so that f ( x ) − L < if x − c < δ1 , and we can find δ2 so that g (x ) − M < 2 2 2 if x − c < δ 2 f (x ) − L < • To make sure both are true, we pick δ = min (δ1 , δ 2 ) • Check: If x − c < δ ≤ δ1 , δ 2 , then f ( x ) + g ( x ) − (L + M ) ≤ f ( x ) − L + g ( x ) − M < Q.E.D. ε ε + =ε 2 2 Corollaries (theorems that follows (almost) immediately) 1) lim ( f ( x ) − g ( x )) = L − M 2) lim (α 1 ⋅ f 1 (x ) + α 2 ⋅ f 2 ( x ) + ... + α n ⋅ f n ( x )) = α 1 L1 + α 2 L2 + ...α n L n if lim f i (x ) = Li 3) lim ( f 1 ( x ) ⋅ f 2 ( x ) ⋅ ... ⋅ f n (x )) = L1 ⋅ L 2 ⋅ ... ⋅ L n x →c x→c x→c x→c 4) If P( x ) = polynomial = a n x n + a n −1 x n −1 + ... + a 0 , then lim P( x ) = P(c ) = a n (c )n + a n −1 (c )n −1 + ... + a 0 x→c Example • lim 2 x 2 + 10 x = 28 x→ 2 Page 16 of 33 MAT137Y1a.doc • Now what happens if the polynomial is in the denominator? Example • lim x→2 1 2 2x + 1 = 1 9 Theorem 2 • If lim g ( x ) = M , M ≠ 0 , then lim x→c x→c 1 1 = g (x ) M • Proof: For ε > 0 we want δ > 0 so that • Algebra: • Choose δ1 so that g (x ) − M < 1 1 − <ε g (x ) M M − g (x ) 1 1 1 − = = g (x ) − M g (x ) M g (x )M g ( x )M M 2 . If x is in the region, then g (x ) − M 1 2 g (x ) − M < = 2 g (x ) − M 2 g ( x )M M M 2 If I choose δ1 so that g (x ) − M < • M2 ε 2 1 1 1 − < g (x ) − M 2 g (x ) M M 2 Check: If x − c < δ , then • < 1 M ⋅ε 2 M 2 2 =ε 2 Lecture #8 – Thursday, October 2, 2003 Examples • • lim 1 2 = 2x + 1 1 lim 3 = −1 x → −1 x x →3 1 19 Theorem 3 • If lim f (x ) = L and lim g (x ) = M , M ≠ 0 , then lim x→c x→c x→c f (x ) L = g (x ) M Page 17 of 33 MAT137Y1a.doc • Proof: lim x→c f (x ) 1 = f (x ) ⋅ g (x ) g (x ) f (x ) 1 = lim f ( x ) ⋅ g (x ) x →c g (x ) = lim f ( x ) ⋅ lim x →c = L⋅ x →c 1 g (x ) 1 L = M M Example • • 2x 2 + x lim 2 x −3 x→ 2 = 10 = 10 1 The hard case: What happens if the denominator approaches 0? Example • lim • lim x →1 1 DNE x −1 x 2 −1 =2 x →1 x − 1 Theorem 4 • If lim f (x ) = L, L ≠ 0 and lim g (x ) = 0 , then lim x→c • x →c Proof: By contradiction • Suppose lim x→c x→c f (x ) =M g (x ) lim f (x ) = lim x→c • Then f (x ) does not exist g (x ) x →c = lim x →c f (x ) ⋅ g (x ) g (x ) f (x ) ⋅ lim g ( x ) g ( x ) x →c = M ⋅0 = 0 • Contradiction because lim f (x ) ≠ 0 . So lim x→c x→c f (x ) DNE. g (x ) Moral If you’re asked to do a limit computation f (x ) , you should proceed in several steps: g (x ) 1) See if the denominator is equal to 0. 2) If NO, plug it in at x = c . Done. If YES, check if numerator limit is 0. 3) If NO, then the limit does not exist. Done. If YES, do algebra to simplify the expression and go back to Step 1. Page 18 of 33 MAT137Y1a.doc Example • • x−2 lim x−2 x→4 lim x→ 4 DNE (by Theorem 4) x2 + 4 =2 x+2 CONTINUITY Idea • A function is continuous if you can draw the graph “without lifting your pencil” Example ( c ) Definition • • Suppose f is defined on (c − p, c + p ) . The function f is continuous at c if: • The limit lim f (x ) exist • the limit lim f (x ) = f (c ) x→ c x→c How can a function not be continuous? f (c ) • Case A: The limit exist, but • Case B: The limit doesn’t exist Example of Case A • c Example of Case B – “Essential Discontinuity” • Different ways it could happen Page 19 of 33 “removable discontinuity” MAT137Y1a.doc • The one-side limits exist, but they don’t agree c • One of the side limits exist, but only one 0, x ≤ 0 f (x ) = 1 sin ,x>0 x • Neither of the one sided limits exist 0, x ∈ Q f (x ) = 1, x ∉ Q Examples • Functions that are continuous: • Polynomials: lim P( x ) = P(c ) x→c • Absolute value (by Building Blocks): lim x = c • Square root: lim x = c for c > 0 x→c x →c Theorem 1 • If f and g are both continuous at c, then f + g , f − g , αf , f ⋅ g , f if g (c ) ≠ 0 are continuous. g Example • h( x ) = x + 3 x + 1 is continuous at any x > 0 x +1 Another Definition For Continuity • f is continuous at c if ∀ε > 0 , ∃δ > 0 such that if x − c < δ then f ( x ) − f (c ) < ε Theorem 2 • If g ( x ) is continuous at c and f is continuous at g (c ) , then f g is continuous at c Page 20 of 33 MAT137Y1a.doc Example g (x ) = x 2 + 1 g(x) = x2 + 1 f (x ) = x f g = x 2 +1 g(x) = x2 + 1 g(c) = 2 f g(1) c=1 • g(1) = 2 Proof: Let ε > 0 . We want to find δ > 0 so that if x − c < δ then f ( g (x )) − f (g (c )) < ε • • Do one function at a time Choose δ1 > 0 so that if t − g (c ) < δ1 , then • Choose δ > 0 so that if x − c < δ , then g (x ) − g (c ) < δ1 because g is continuous at c • g (c ) f (t ) − f (g (c )) < ε because f is continuous at Check: If x − c < δ then g (x ) − g (c ) < δ1 then f ( g (x )) − f (g (c )) < ε Example 1 (continuous, c ≠ 0 ) x • If f (x) = x , g (x ) = x 2 + 1 , h( x ) = • k (x ) = h g f (x ) 1 is also continuous for any c = 2 x +1 Detecting Discontinuities • Definition: f is continuous from the left if lim− f (x ) = f (c ) ; f is continuous from the right if lim+ f (x ) = f (c ) x →c x →c x=c Another Way To Say A Function Is Continuous • f is continuous at x = c iff • lim− f ( x ) exists and is equal to f (c ) x→c • lim f (x ) exists and is equal to f (c ) x→c + Page 21 of 33 MAT137Y1a.doc Example Determine the discontinuities of the function: 2 x − 1 if x < 1 g ( x ) = 0 if x = 1 1 x2 • • if x > 1 Certainly the function is continuous at any point in (− ∞,1) or any point in (1, ∞ ) . So the only point in question is c = 1 . 1 lim g (x ) = lim= 2 x − 1 = 1 and lim+ g (x ) = lim+ 2 = 1 which is not equal to g (1) = 0 . So this function is x →1= x →1 x →1 x →1 x not continuous. c = 1 is removable discontinuity. Last Definition (For Now) For Continuity Of Fuctions • Let (a, b ) be an interval where f is defined. • Definition: f is continuous on (a, b ) if f is continuous at every point c in • • Let [a, b ] be an interval where f is defined. Definition: f is continuous on [a, b ] if f is continuous on (a, b ) and it’s continuous from the right at a and continuous from the left at b (a, b ) a c b Lecture #9 – Tuesday, October 7, 2003 THEOREM “THE PINCHING THEOREM” • Let p > 0 . Suppose that for all x such that • If lim h( x ) = L = lim g (x ) , then lim f (x ) = L 0 < x − c < p , we have h(x ) ≤ f (x ) ≤ g (x ) . x→c x →c x→c c-p • Proof: Let ε > 0 . Want: • • • c c+p f ( x ) − L < ε for x − c small enough. Choose δ1 so that h(x ) − L < ε if x − c < δ1 Choose δ2 so that g (x ) − L < ε if x − c < δ 2 L − ε < h( x ) < L + ε L − ε < g (x ) < L + ε If we choose δ = min (δ1 , δ 2 ) , then L − ε < h( x ) ≤ f ( x ) ≤ g ( x ) < L + ε Q.E.D. Page 22 of 33 f (x ) − L < ε MAT137Y1a.doc APPLICATION: SINE AND COSINE First A Figure Q P 1 x tan x sin x cos x O B A 1 • First task: Compute lim sin x = 0 (1) x→0 • Proof: From the figure, for x small enough sin x < x • • So in fact, − x < sin < x for x small So by Pinching Theorem, lim (− x ) = 0 and lim (x ) = 0 , so lim sin x = 0 x →c x→c x→0 Q.E.D. • • Because (by Pythagorean Theorem) sin 2 x + cos 2 x = 1 , cos x = 1 − sin 2 x for x small So because of the theorem about composing continuous functions, ( ) lim cos x = lim 1 − sin 2 x = lim 1 − sin 2 x = 1 (because lim sin x = 0 ) x→0 x →0 x→0 • Therefore, lim cos x = 1 (2) • Now what about x→0 x→0 lim sin x = sin c x→c lim cos x = cos c sine and cosine are continuous x→c • Proof of (3): • lim sin x is the same as lim sin (c + h ) x→c h →0 • Now, we can use the addition formulas: sin (c + h ) = sin c cosh + cos c sinh • h →0 lim sin (c + h ) = (sin c ) lim cosh + (cos c ) lim sinh h →0 = sin c h →0 Q.E.D. Lecture #10 – Thursday, October 9, 2003 • In fact, all the trigonometric functions are continuous (where defined): this follows from the theorem about quotients of continuous functions being continuous. Page 23 of 33 MAT137Y1a.doc TWO MORE IMPORTANT LIMITS • • lim x→0 sin x 1 − cos x = 1 (1) and lim = 0 (2) x→0 x x Note: For both of these, if you “plug in”, you get “ 0 ” – need a subtler argument 0 Proof of (1) • Refer to Figure • Area of ∆OBP = • • • • 1 (sin x )(1) = sin x 2 2 x Area of sector OAP = 2 tan x Area of ∆OAQ = 2 sin x x tan x < < 2 2 2 sin x sin x 1 <1< ⋅ x x cos x sin x 1 1< ⋅ sin x x cos x Two inequalities: < 1 and sin x x cos x < x sin x < x < sin x cos x sin x < 1 – true for x “small” and x > 0 or x < 0 x • So cos x < • By Pinching Theorem, we can conclude that lim sin x = 1 because lim cos x = 1 and lim 1 = 1 . x→0 x →0 x→0 x Q.E.D. Proof of (2) • • • 1 − cos x 1 − cos x 1 + cos x 1 − cos 2 x sin 2 x sin x sin x = ⋅ = = = ⋅ x x 1 + cos x x(1 + cos x ) x(1 + cos x ) x (1 + cos x ) 1 − cos x sin x sin x Therefore lim = lim ⋅ lim =0 x→0 x →0 x x →0 (1 + cos x ) x Q.E.D. sin (ax ) 1 − cos(ax ) = 0 and lim =0 x→0 x→0 ax ax In fact, lim Examples of Application 1) 2) lim x→0 sin (5 x ) sin (5 x ) 5 sin (5 x ) 5 5 sin (5 x ) 5 = lim ⋅ = lim ⋅ = lim = x →0 4x 4x 5 x →0 5 x 4 4 x →0 5 x 4 2x 2 + x x(2 x + 1) x = lim = lim ⋅ lim (2 x + 1) = x → 0 sin x x →0 sin x x →0 sin x x →0 lim 1 x x →0 sin x lim Page 24 of 33 ⋅ lim (2 x + 1) = 1 x →0 MAT137Y1a.doc x2 = lim x → 0 sec x − 1 x →0 lim 3) = lim x 2 cos x(1 + cos x ) x2 x2 x 2 cos x = lim = lim = lim 1 x →0 1 − cos x x →0 1 − cos x x →0 (1 − cos x )(1 + cos x ) −1 cos x cos x x 2 cos x + x 2 cos 2 x 2 x →0 = lim x →0 x→0 sin x x sin x = lim 2 cos x + lim x →0 x sin x x2 2 sin x 2 ⋅ cos x + lim x →0 x2 2 sin x ⋅ cos 2 x cos 2 x = 2 TWO VERY IMPORTANT THEOREMS ABOUT CONTINUOUS FUNCTIONS • • • Intermediate Value Theorem Extreme Value Theorem Understanding of the proof is going to require the Least Upper Bound Axioms Idea For Intermediate Value Theorem (IVT) • A continuous function has a graph which is “an unbroken curve” f(b) f(b) k f(a) f(a) a b a continuous b non-continuous THEOREM: INTERMEDIATE VALUE THEOREM • If f is continuous on [a, b ] and k is any number between f (a ) and f (b ) , then there is a value c, a < c < b so that f (c ) = k Application • Locating the zeros of a function • Suppose f so continuous on [a, b ] and f (a ) < 0 and f (b ) > 0 , then there is a solution c to f (c ) = 0 solution between a and b (by IVT) a b Page 25 of 33 MAT137Y1a.doc Example f (x ) = cos π x − x 2 on [0,1] 2 • Evaluate at x = 0 : cos π ⋅ 0 − (0 )2 = 1 > 0 2 • Evaluate at x = 1 : cos π ⋅1 − (1)2 = −1 < 0 2 • IVT says there is a solution to cos • Now evaluate at x = • Now IVT says there is root between • Can keep going and get an approximation to root – “bisection method” π x − x 2 = 0 between [0,1] 2 1 1 π 1 : f = cos − 2 4 2 2 2 ≈ 0.45 > 0 1 and 1. 2 Application: Solving Inequalities • We were implicitly using IVT before! • Solve the inequality x( x + 2)( x − 3) > 0 • Find the zeros: 0, -2, 3 -2 • 0 3 If a, b are 2 zeros of P(x ) = x( x + 2)(x − 3) , then the IVT is saying that on (a, b ) , P(x ) has to be all positive or negative – because if it’s positive f (a ) > 0, a ∈ (a, b ) , f (b ) < 0, a ∈ (a, b ) then there is a zero between a and b; that’s a contradiction since you’ve already found all the zeros BOUNDEDNESS AND EXTREME VALUES • • Let f be a function, defined on an interval I Definition: f is called bounded on I if there are constants k and K so that k < f ( x ) < K for all x ∈ I Example • f (x ) = x 2 on [0,2] 4 2 Page 26 of 33 MAT137Y1a.doc • f (x ) = sin x on [0,2π] 1 2π -1 • Definition: If f is not bounded it is called unbounded (on I) Example • 1 on (0, ∞ ) g (x ) = x 2 0 at x = 0 • • • g ( x ) is unbounded on (0, ∞ ) g ( x ) is bounded on [1, ∞ ) Note: A function can have a maximum value or a minimum value or both or neither on an interval I Example • 1 on (0, ∞ ) g (x ) = x 2 0 at x = 0 • • • • On [0, ∞ ) : unbounded, no maximum, has minimum On [1, ∞ ) : bounded, has maximum, no minimum On (1, ∞ ) : bounded, no maximum, no minimum Note: This has nothing to do with continuity Example • f (x ) = 1 if x ∈ Q − 1 if x ∉ Q is bounded, has maximum, has minimum on (− ∞, ∞ ) THEOREM: EXTREME VALUE THEOREM • If a function is continuous on a bounded, closed interval [a, b ] , then the function takes on both a maximum value M and a minimum value m – i.e. m ≤ f ( x ) ≤ M on [a, b ] (m and M are called the “extreme values”) Page 27 of 33 MAT137Y1a.doc Lecture #11 – Tuesday, October 14, 2003 • Warning: Need all three of the assumptions Counter-Examples 1) The interval must be bounded f (x ) = x 2 on [0, ∞ ) 2) The interval must be closed f (x ) = x 2 on (0,1) 3) The function must be continuous a b Idea of IVT • A continuous function “takes intervals to intervals” Idea of IVT and EVT Together • A continuous function “takes bounded closed intervals to bounded closed intervals” • Proofs of IVT and EVT require an understanding of the “Least Upper Bound Axiom” of the real number LEAST UPPER BOUNDS • Let S be a nonempty set of real numbers Example • S1 = (− ∞,0 ) • T1 = [0, ∞ ) • T2 = [1, ∞ ) • T1 = φ 0 1 2 n , ,..., ,... 2 3 n +1 • S2 = • S 3 = (1,2,3,.., n,...) 0 1 x 1 2 3 4 5 • S4 = • Definition: A number M is an upper bound for S if x ≤ M for any x ∈ S (x + 1)(x + 3) :3≤ x ≤ 5 Page 28 of 33 MAT137Y1a.doc • • • • Note: Not all sets have upper bounds – those with upper bounds in the example are S1, S2, S4 Definition: If S has an upper bound, we say S is bounded above Now let’s think about the set of possible upper bounds for S (call it T) What is the smallest possible upper bound? – in the Example, for S1, the smallest is 0; for S2, the smallest is 1 Definition • If S is nonempty set of real numbers that is bounded above, the least upper bound (lub) of S is an upper bound that is less than or equal to any upper bound for S LEAST UPPER BOUND AXIOM • Every nonempty set of real numbers that has an upper bound has a least upper bound • If you think this is “obvious”, it’s NOT TRUE, for example, the rational numbers! Example { } • S = x < 2 , x is rational • Let T = the set of possible rational upper bound • for S = x > 2 : x is rational Has no least “element” { 0 2 } Back To Real Numbers: Examples 1) lub(− 4,−1) = −1 2) 1 1 1 lub − 1,− ,− ,...,− 3 ,... = 0 8 27 n 3) lub x ∈ R : x 2 < 3 = 3 { } Lecture #12 – Thursday, October 16, 2003 Theorem • If M is the least upper bound of set S and ε > 0 is any positive number, then there is an element s in S such that M − ε < s ≤ M elements in S M-ε • M Idea: The elements in S get arbitrarily close to M Proof • • • Let ε > 0 . Then since M is an upper bound, any element x of S satisfy x < M So we only need to find s ∈ S so that M − ε < s We’ll argue by contradiction. Suppose for all s ∈ S , s ≤ M − ε • But that means M − ε is an upper bound for S • But M − ε < M and M is supposed to be the lub: contradiction Page 29 of 33 MAT137Y1a.doc • So there must be s, M − ε < s ≤ M Example S= 1 2 n , ,..., ,... 2 3 n +1 999/1000 0 • Take ε = 1/2 1 1 999 1000 9999 . Is it true that there is an element of S such that < s < 1 ? Yes: s = , 1000 1000 1001 10000 LOWER BOUND Definition • Let S be a nonempty set of real numbers, then we say m is a lower bound of S if ∀x ∈ S , m ≤ x – if S has a lower bound, then S is bounded below • If S is a nonempty set of real numbers bounded below, then the greatest lower bound (glb) is a lower bound for S that is greater than or equal to any other lower bound for S Theorem: Existence of Greatest Lower Bound • Every nonempty set of S of R bounded below has a greatest lower bound Theorem • If m is the glb of S, ε > 0 , then ∃ an element s ∈ S such that m ≤ s < m + ε Proofs • Use what you know for least upper bounds PROOF OF INTERMEDIATE VALUE THEOREM Lemma • Let f be continuous on [a, b ] . If f (a ) < 0 < f (b ) (or f (a ) > 0 > f (b ) ) then ∃ c, a < c < b such that f (c ) = 0 Page 30 of 33 MAT137Y1a.doc Picture ξ c a b f(a) Proof • • • • • Suppose f (a ) < 0 < f (b ) (The other situation can be proved similarly) Let S = {ξ : f is negative on [a, ξ )} By continuity of f, S is not empty Because f (b ) > 0 , b is an upper bound for S. That means S has a least upper bound: call it c Want: f (c ) = 0 (we’ll argue that it can’t be >0 and can’t be <0) • • Suppose f (c ) > 0 • By continuity of f at c, there is an interval (c − ε, c + ε ) where f is >0 • Then because c − ε > 0 , c− ε is also an upper bound for S, but c is the 2 least upper bound. Contradiction. Suppose f (c ) > 0 • By continuity of f at c, f is <0 on an interval (c − ε, c + ε ) ε is in S, but c is an upper bound 2 ε • Contradiction because c + > c 2 So f (c ) is neither <0 or >0, so f (c ) = 0 Q.E.D. • • f So c + Theorem: IVT • If f is continuous on [a, b ] , and k is any value between f (a ) and f (b ) , then ∃ c, a < c < b , such that f (c ) = k Proof • • • • • Suppose f (a ) < k < f (b ) (The other case is similar) Consider g ( x ) = f (x ) − k Then g (a ) < 0 , g (b ) > 0 so g is continuous on [a, b ] By the Lemma, there is a c, a < c < b so that g (c ) = 0 But then if g (c ) = 0 = f (c ) − k , then f (c ) = k Q.E.D. Page 31 of 33 MAT137Y1a.doc PROOF OF EXTREME VALUE THEOREM Lemma • If f is continuous on [a, b ] , then f is bounded on [a, b ] Proof • • Let S = {x ∈ R : x ∈ [a, b] and f is bounded on [a, x ]} • S is bounded above by definition of S: b is an upper bound • S is not empty: can take x = a Let c = lub S . Want: c = b • Because b is an upper bound and c is the least upper bound, then c ≤ b . • We must rule out the possibility that c < b . • Suppose c < b • By continuity of f at c, f is bounded on some interval (c − ε, c + ε ) . • ε . f is also bounded on 2 c− ε ε ,c+ 2 2 ε ε is in S. c+ 2 2 • But c is an upper bound for S, so contradiction. • So c = b . Want: f is bounded on [a, b ] . • Because f is bounded at b, there is an interval [b − δ, b ] where f is bounded. is bounded on • a, c − Because c is the lub, f is bounded on a, c + • Because b is lub of S, f is bounded on • So f is bounded on [a, b ] . a, b − ε . 2 Theorem: EVT • If f is continuous on [a, b ] , then f achieves both a maximum M and minimum m value on [a, b ] . M a c b m Proof • • • By Lemma, f is bounded. Definition: M = lub{ f ( x ) : x ∈ [a, b]} Want: ∃ c such that f (c ) = M . • Suppose not. Then f (x ) ≠ M for all x ∈ [a, b ] . • Then M − f (x ) ≠ 0 for all x ∈ [a, b ] (in fact M − f ( x ) > 0 for all x ∈ [a, b ] ). • Define g ( x ) = • Because M is lub, M − f (x ) can be made arbitrarily small, and g ( x ) can be made arbitrarily large. 1 . So g is continuous on [a, b ] . M − f (x ) Page 32 of 33 f MAT137Y1a.doc • • So g is continuous on [a, b ] , but not bounded. This contradicts the Lemma. Theforefore, there is a c such that f (c ) = M . Q.E.D. Page 33 of 33

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