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Limit of a Function……………………………………………..……………1 Calculus Limits Limit of a function……………………………………………..……………1 One-Sided Limits……………………………………………….…..………..1 Infinite limits…………………………………………………………………2 Vertical Asymptotes.........................................................................................3 Calculating Limits Using the Limit Laws…………… …………………….5 The Squeeze Theorem ………………………………… ………………….6 The Precise Definition of a Limit…………………..…….……………….....7 Continuity…………………………………………………………………….8 Intermediate Value Theorem……….……………………………………….9 References……………………………………………………………………..…9 0 Calculus Limits Limit of a function Lim f() x= L if we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close x−> a to a but not equal to a. t 2 +9 − 3 Example: Let f(x) = . Discuss the behavior of the values of f(x) when x is close to 0. t 2 Solution: Make a table to see the behavior… t t 2 +9 − 3 t 2 ± 0.1 0.16228 ± 5.0 0.16553 ± 1.0 0.16662 ± 0.05 0.16666 As t approaches 0, the values of the function seem to approach 0.16666… t 2 +9 − 3 lim = 0.16666 t→0 t 2 One-Sided Limits Definition: We write Lim f() x= L and say the left-hand limit of f(x) as x approaches a is equal to L if x→ a− we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close to a and less than a i.e x approaches a from the left. 1 Calculus Limits Definition: We write Lim f() x= L and say the right-hand limit of f(x) as x approaches a is equal to L x→ a+ if we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close to a and greater than a i.e x approaches a from the right Lim f() x= L if and only if Lim f() x= L and Lim f() x= L x→ a x→ a− x→ a+ Infinite limits Definition: Let f be a function defined on both sides of a. Then Lim f() x = ∞ means that the values of f(x) can be made arbitrarily large by taking x sufficiently close x→ a to a, but not equal to a. 2 Calculus Limits Definition: Let f be a function defined on both sides of a. Then Lim f() x = −∞ means that the values of f(x) can be made arbitrarily large negative by taking x x→ a sufficiently close to a, but not equal to a. Vertical Asymptotes Definition: The line x = a is called a vertical asymptote of the curve y = f(x) if at least one of the following statements is true: Lim f() x = ∞ Lim f() x = ∞ Lim f() x = ∞ x→ a x→ a− x→ a+ Lim f() x = −∞ Lim f() x = −∞ Lim f() x = −∞ x→ a x→ a− x→ a+ 3 Calculus Limits Example: Find the vertical asymptotes of f(x) = tanx sin x Solution: Because tan x = cos x There are potential vertical asymptotes where cosx = 0. In fact, since cosx → 0+ as x → (π )− and cosx → 0− as x → (π )+ , whereas sin x is positive when x is near π , we have 2 2 2 lim tan x = ∞ and lim tan x = −∞ . This shows that the line x = π is a vertical asymptote. x→( π x→(π 2 )2 − )2 + Similar reasoning shows that the lines x=(2 n + 1)π , (odd multiples of π ),where n is an integer, 2 2 are all vertical asymptotes of f(x) = tanx. 4 Calculus Limits Calculating Limits Using the Limit Laws Suppose that c is a constant and the limits Lim f() x and Lim g() x exist. Then x→ a x→ a 1. Lim[ f ( x )+ g ( x )] = lim f ( x ) + lim g ( x ) x→ a x→ a x→ a 2. Lim[ f ( x )− g ( x )] = lim f ( x ) − lim g ( x ) x→ a x→ a x→ a 3. Lim[ cf ( x )]= c lim f ( x ) x→ a x→ a 4. Lim[ f ( x ) g ( x )]= lim f ( x ).lim g ( x ) x→ a x→ a x→ a f() x limf ( x ) 5. Lim[ ] = x→ a iflim g ( x )≠ 0 x→ a g() x limg ( x ) x→ a x→ a 6. Lim[ f ( x )]n = [lim f ( x )]n where n is a positive integer. x→ a x→ a 7. Lim c= c 8. Lim x= a x→ a x→ a 9. Lim xn= a n , where n is a positive integer. x→ a 10. Limn x= n a , where n is a positive integer. x→ a 11. Limn f( x )= n lim f ( x ) , where n is a positive integer. x→ a x→ a x3+2 x 2 − 1 Evaluate lim . x→−2 5− 3x 5 Calculus Limits 3 2 x3+2 x 2 − 1 lim(x+ 2 x − 1) Solution: lim = x→−2 x→−2 5− 3x lim(5− 3x ) x→−2 limx3 + 2 lim x 2 − lim1 = x→−2 x→−2 x→−2 lim 5− 3 lim x x→−2 x→−2 (− 2)3 + 2( − 2)2 − 1 = 5− 3( − 2) 1 = − 11 x 2 −1 Example. Find lim x→1 x −1 x 2 −1 Solution: Let f(x) = . We can’t find the limit by substituting x=1 because f(1) isn’t defined. Nor can x −1 we apply the quotient rule because the limit of the denominator is 0. Instead we factor the numerator as a difference of squares: x 2 −1 (x− 1)( x + 1) = . The numerator and denominator have a common factor of x-1. When we take x −1 x −1 the limit as x approaches 1, we have x ≠1 so x −1≠ 0 . Therefore, we can cancel the common factor and compute the limit as follows x 2 −1 (x− 1)( x + 1) lim = lim x→1 x −1 x→1 x −1 = lim(x + 1) x→1 = 1 +1 = 2 The Squeeze Theorem If f()()() x≤ g x ≤ h x when x is near a and Lim f() x = Lim h() x= L then x→ a x→ a Lim g() x= L . x→ a 6 Calculus Limits 1 Example. Show that limx 2 sin = 0 . x→0 x 1 1 1 Solution: First note that we cannot use limx 2 sin = 0 = limx 2 .limsin because limsin does x→0 x x→0 x→0 x x→0 x 1 not exist. However since −1 ≤ sin ≤1, x Multiply all sides by x 2 1 −x2 ≤ x 2 sin ≤ x 2 . x We know that limx 2 = 0 and lim(−x 2 ) = 0 x→0 x→0 1 Taking f( x )= − x2 , g ( x ) = x 2 sin( ),h ( x ) = x 2 in the squeeze theorem, we obtain x 1 limx 2 sin = 0 x→0 x The Precise Definition of a Limit Let f be a function on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f(x) as x approaches a is L, and we write Lim f() x= L if for every number ε > 0 there is a number δ > 0 such that x→ a f() x− L < ε whenever 0 <x − a < δ . Example prove that lim(4x − 5) = 7 . x→3 Solution 1. Preliminary analysis of the problem (guessing a value for δ ). Let ε be a given positive number. We want to find a number δ such that (4x − 5) − 7 < ε whenever 0<x − 3 < δ 7 Calculus Limits But (4x − 5) − 7 = 4x − 12 = 4( x − 3) = 4 x − 3 . Therefore, we want 4x − 3 < ε whenever ε 0<x − 3 < δ that is, x −3 < whenever 0<x − 3 < δ . This suggests that we should choose 4 δ = ε . 4 2. Proof (showing that this δ works). Given ε > 0 , choose δ = ε . If , then 4 0<x − 3 < δ ε (4x −−=−= 5) 7 4x 12 4( x −=−<= 3) 4 x 3 4δ 4( ) = ε 4 Thus (4x − 5) − 7 < ε whenever 0<x − 3 < δ Therefore by definition of a limit lim(4x − 5) = 7 . x→3 Continuity A function f is continuous at a number a if limf ( x )= f ( a ) x→ a Notice that the definition implicitly requires three things if f is continuous at a: 1. f(a) is defined (that is, a is in the domain of f) 2. limf ( x ) exists. x→ a 3. limf ( x )= f ( a ) x→ a A function f is continuous on an interval if it is continuous at every number in the interval. Example: Show that the function f( x )= 1 − 1 − x 2 is continuous on the interval [-1,1]. Solution: If -1 < a < 1, then using the Limit Laws, we have Lim f( x )= lim(1 − 1 − x 2 ) x→ a x→ a 8 Calculus Limits = 1− lim 1 − x 2 x→ a = 1− lim(1 − x 2 ) x→ a = 1− 1 − a 2 = f(a) Therefore f is continuous on [-1.1]. If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: 1. f + g 2. f – g 3. cf 4 fg 5. f/g if g(a) ≠ 0 Any polynomial is continuous everywhere, that is it continuous on ℜ = (,)−∞ ∞ Any rational function is continuous wherever it is defined, that is, it is continuous on its domain. Intermediate Value Theorem Suppose that f is continuous on the closed interval [a,b] and let N be any number between f(a) and f(b), where f( a )≠ f ( b ). Then there exists a number c in (a,b) such that f(c) = N.
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