MATH 12002 - CALCULUS I 1.6: Infinite Limits Professor Donald L. - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §1.6: Infinite Limits

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 9

Introduction to Infinite Limits

Our definition of lim

x→a f (x) = L required a and L to be real numbers.

In this section, we expand the definition to allow a to be infinite (limits at infinity) or L to be infinite (infinite limits). We first consider infinite limits. A function y = f (x) has an infinite limit at x = a if the values of y become arbitrarily large (either positive or negative) when x is close to a. Our basic definition is:

Definition

Let y = f (x) be a function and let a be a number. The limit of f as x approaches a is +∞ if y can be made arbitrarily large (and positive) by taking x close enough to a, but not equal to a. We write lim

x→a f (x) = +∞.

Compare this to the definition of lim

x→a f (x) = L.

D.L. White (Kent State University) 2 / 9

Introduction to Infinite Limits

Most of the functions we study that have infinite limits at a point x = a are quotients of functions. To evaluate these infinite limits, we will use the following idea.

Basic Principle

Let f (x) = N(x)/D(x). If lim

x→a D(x) = 0 and lim x→a N(x) = 0, then the

• ne-sided limits at x = a of f (x) are infinite; that is

lim

x→a−

N(x) D(x) = ±∞ and lim

x→a+

N(x) D(x) = ±∞. If the signs are the same, then lim

x→a N(x) D(x) = ±∞.

If the signs are different, then lim

x→a N(x) D(x) does not exist.

D.L. White (Kent State University) 3 / 9

Introduction to Infinite Limits

The Basic Principle says that in order to find lim

x→af (x),

where f (x) = N(x)

D(x) with lim x→a D(x) = 0 and lim x→a N(x) = 0,

we only need to determine the signs of lim

x→a−f (x) and lim x→a+f (x). That is,

1 Determine the sign of y = f (x) = N(x)

D(x) as x → a−.

If +, then lim

x→a−f (x) = +∞.

If −, then lim

x→a−f (x) = −∞.

2 Determine the sign of y = f (x) = N(x)

D(x) as x → a+.

If +, then lim

x→a+f (x) = +∞.

If −, then lim

x→a+f (x) = −∞.

3 If lim

x→a−f (x) = lim x→a+f (x) = ±∞, then lim x→af (x) = ±∞.

If lim

x→a−f (x) = lim x→a+f (x), then lim x→af (x) does not exist.

D.L. White (Kent State University) 4 / 9

Examples

Example

Find lim

x→3

x2 + 7 x2 − 8x + 15 if the limit exists or is ±∞.

Solution

First notice that as x → 3, x2 + 7 → 32 + 7 = 16 = 0 and x2 − 8x + 15 → 32 − 8(3) + 15 = 9 − 24 + 15 = 0, so the one-sided limits are infinite. [Continued →]

D.L. White (Kent State University) 5 / 9

Examples

Solution [continued]

Notice also that x2 − 8x + 15 = (x − 3)(x − 5), so f (x) =

x2+7 (x−3)(x−5).

1

lim

x→3− f (x) : As x → 3−

x2 + 7 → 16 x − 5 → −2 x − 3 → 0−, and so f (x) =

x2+7 (x−3)(x−5) is positive. Hence

lim

x→3−f (x) = +∞.

[Continued →]

D.L. White (Kent State University) 6 / 9

Examples

Solution [continued]

2

lim

x→3+ f (x) : As x → 3+

x2 + 7 → 16 x − 5 → −2 x − 3 → 0+, and so f (x) =

x2+7 (x−3)(x−5) is negative. Hence

lim

x→3+f (x) = −∞.

3 Since lim

x→3−f (x) = lim x→3+f (x),

lim

x→3f (x) DOES NOT EXIST.

D.L. White (Kent State University) 7 / 9

Examples

Example

Find lim

x→2

x3 − 9 x2 − 4x + 4 if the limit exists or is ±∞.

Solution

First notice that as x → 2, x3 − 9 → 23 − 9 = −1 = 0 and x2 − 4x + 4 → 22 − 4(2) + 4 = 4 − 8 + 4 = 0, so the one-sided limits are infinite. [Continued →]

D.L. White (Kent State University) 8 / 9

Examples

Solution [continued]

Notice also that x2 − 4x + 4 = (x − 2)2, so f (x) =

x3−9 (x−2)2 .

In contrast to the previous example, as x → 2, we have (x − 2)2 → 0, but (x − 2)2 is positive whether x > 2 or x < 2. Therefore, as x → 2 (from either side), x3 − 9 → −1 (x − 2)2 → 0+, and so f (x) =

x3−9 (x−2)2 is negative. Hence

lim

x→2f (x) = −∞.

D.L. White (Kent State University) 9 / 9