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

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Calculus Limits

Limit of a function )( = LxfLim if we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close >− ax to a but not equal to a.

t 2 −+ 39 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 −+ 39

t 2 ± 0.1 0.16228

± 5.0 0.16553 ± 1.0 0.16662 ± 05.0 0.16666 As t approaches 0, the values of the function seem to approach 0.16666… t 2 −+ 39 lim = 0.16666 t→0 t 2

One-Sided Limits Definition: We write )( = LxfLim and say the left-hand limit of f(x) as x approaches a is equal to L if →ax − 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.

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Calculus Limits

Definition: We write )( = LxfLim and say the right-hand limit of f(x) as x approaches a is equal to L →ax + 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

)( = LxfLim if and only if )( = LxfLim and )( = LxfLim →ax →ax − →ax +

Infinite limits

Definition: Let f be a function defined on both sides of a. Then xfLim )( ∞= means that the values of f(x) can be made arbitrarily large by taking x sufficiently close →ax to a, but not equal to a.

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Calculus Limits

Definition: Let f be a function defined on both sides of a. Then xfLim )( −∞= means that the values of f(x) can be made arbitrarily large negative by taking x →ax 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: xfLim )( ∞= xfLim )( ∞= xfLim )( = ∞ →ax →ax − →ax + xfLim )( −∞= xfLim )( −∞= xfLim )( = −∞ →ax →ax − →ax +

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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 x → 0cos + as x → (π )− and x → 0cos − as x → (π )+ , whereas sin x is positive when x is near π , we have 2 2 2 tanlim x ∞= and tanlim 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 nx += )12( π , (odd multiples of π ),where n is an integer, 2 2 are all vertical asymptotes of f(x) = tanx.

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Calculus Limits

Calculating Limits Using the Limit Laws

Suppose that c is a constant and the limits xfLim )( and xgLim )( exist. Then →ax →ax

1. +=+ xgxfxgxfLim )(lim)(lim)]()([ →ax →ax →ax 2. −=− xgxfxgxfLim )(lim)(lim)]()([ →ax →ax →ax 3. = xfcxcfLim )(lim)]([ →ax →ax

4. = xgxfxgxfLim )(lim).(lim)]()([ →ax →ax →ax xf )( xf )(lim 5. Lim[ ] = →ax xgif ≠ 0)(lim →ax xg )( xg )(lim →ax →ax 6. n = xfxfLim )](lim[)]([ n where n is a positive integer. →ax →ax 7. = ccLim 8. = axLim →ax →ax 9. = axLim nn , where n is a positive integer. →ax 10. n = n axLim , where n is a positive integer. →ax

11. n = n xfxfLim )(lim)( , where n is a positive integer. →ax →ax

xx 23 −+ 12 Evaluate lim . x −→ 2 − 35 x

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Calculus Limits

23 xx 23 −+ 12 xx −+ )12(lim Solution: lim = x −→ 2 x −→ 2 − 35 x − x)35(lim x −→ 2 3 xx 2 −+ 1limlim2lim = x −→ 2 x −→ 2 x −→ 2 − lim35lim x x −→ 2 x −→ 2 3 2 −−+− 1)2(2)2( = −− )2(35 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 xx +− )1)(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 − ≠ 01 . Therefore, we can cancel the common factor and compute the limit as follows x 2 −1 xx +− )1)(1( lim = lim x→1 x −1 x→1 x −1 = x + )1(lim x→1 = 1 +1 = 2

The Squeeze Theorem

If ≤≤ xhxgxf )()()( when x is near a and xfLim )( = )( = LxhLim then →ax →ax )( = LxgLim . →ax

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Calculus Limits

1 Example. Show that x 2 sinlim = 0 . x→0 x 1 1 1 Solution: First note that we cannot use x 2 sinlim = 0 = x 2 sinlim.lim because sinlim does x→0 x x→0 x→0 x x→0 x 1 not exist. However since sin1 ≤≤− 1, x Multiply all sides by x 2 1 xx 22 sin ≤≤− x 2 . x We know that x 2 = 0lim and x 2 =− 0)(lim x→0 x→0 1 Taking 2 =−= xxgxxf 2 sin()(,)( )(), = xxh 2 in the squeeze theorem, we obtain x 1 x 2 sinlim = 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

)( = LxfLim if for every number ε > 0 there is a number δ > 0 such that →ax )( Lxf <− ε whenever 0 ax <−< δ .

Example prove that x =− 7)54(lim . 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 x 7)54( <−− ε whenever x 30 <−< δ

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Calculus Limits

But x xxx −=−=−=−− 34)3(41247)54( . Therefore, we want x 34 <− ε whenever ε x 30 <−< δ that is, x 3 <− whenever x 30 <−< 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 x 30 <−< δ ε x xxx δ (4434)3(41247)54( ) ==<−=−=−=−− ε 4 Thus x 7)54( <−− ε whenever x 30 <−< δ Therefore by definition of a limit x =− 7)54(lim . x→3 Continuity

A function f is continuous at a number a if = afxf )()(lim →ax

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. xf )(lim exists. →ax 3. = afxf )()(lim →ax

A function f is continuous on an interval if it is continuous at every number in the interval. Example: Show that the function xf 11)( −−= x 2 is continuous on the interval [-1,1]. Solution: If -1 < a < 1, then using the Limit Laws, we have xfLim −−= x 2 )11(lim)( →ax →ax

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Calculus Limits

= 1lim1 −− x 2 →ax = −− x 2 )1(lim1 →ax = 11 −− 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 ≠ bfaf ).()( Then there exists a number c in (a,b) such that f(c) = N.

Example. Using the Intermediate Value Theorem, let’s Show that there is a root of the equation 23 xxx =−+− 02364 between 1 and 2. Solution: let f(x) = 23 xxx −+− 2364 . We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c)= 0. Therefore, we take a=1, b=2, N=0. We have f(1) = 4 – 6 + 3 – 2 = -1 < 0 f(2) = 32 – 24 + 6 – 2 = 12 > 0 Thus, f(1) < 0 < f(2); that is, N = 0 is a number between f(1) and f(2). Now f is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0. In other words, the equation 23 xxx =−+− 02364 has at least one root c in the interval (1,2).

References: Calculus, James Stewart 5th edition

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