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

MATH 12002 - CALCULUS I §1.6: Limits at Infinity

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 13

Introduction to Limits at Infinity

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 now consider limits at infinity. A function y = f (x) has limit L at infinity if the values of y become arbitrarily close to L when x becomes large enough. Our basic definition is:

Definition

Let y = f (x) be a function and let L be a number. The limit of f as x approaches +∞ is L if y can be made arbitrarily close to L by taking x large enough (and positive). We write lim

x→+∞ f (x) = L.

Compare this to the definition of lim

x→a f (x) = L.

The definition means that the graph of f is very close to the horizontal line y = L for large values of x.

D.L. White (Kent State University) 2 / 13

Introduction to Limits at Infinity

Most of the functions we study that have finite limits at infinity are quotients of functions. To evaluate these limits at infinity, we will use the following idea.

Basic Principle

If c is a real number and r is any positive rational number, then lim

x→+∞

c xr = 0. If c is a real number and r is any positive rational number such that xr is defined for x < 0, then lim

x→−∞

c xr = 0. This is used as in the following examples.

D.L. White (Kent State University) 3 / 13

Examples

Example

Find lim

x→+∞

2x3 + 5 7x3 + 4x + 3 if the limit exists. Before we solve this problem, notice that 2x3 + 5 and 7x3 + 4x + 3 both approach +∞ as x → +∞ and so lim

x→+∞

2x3 + 5 7x3 + 4x + 3 = lim

x→+∞(2x3 + 5)

lim

x→+∞(7x3 + 4x + 3).

In order to evaluate the limit, we will first use an algebraic manipulation to turn this into an expression whose limit is the quotient of the limits of the numerator and denominator. We will then use the Basic Principle to evaluate these limits.

D.L. White (Kent State University) 4 / 13

Examples

Solution

Divide the numerator and denominator by the highest power of x that appears in the denominator; in this case x3. That is, lim

x→+∞

2x3 + 5 7x3 + 4x + 3 = lim

x→+∞

(2x3 + 5) 1

x3

(7x3 + 4x + 3) 1

x3

= lim

x→+∞ 2x3 x3 + 5 x3 7x3 x3 + 4x x3 + 3 x3

= lim

x→+∞

2 + 5

x3

7 + 4

x2 + 3 x3

. [Continued →]

D.L. White (Kent State University) 5 / 13

Examples

Solution [continued]

Now the Basic Principle says that

5 x3 , 4 x2 , and 3 x3 all approach 0 as

x → ∞, and so lim

x→+∞

2x3 + 5 7x3 + 4x + 3 = lim

x→+∞

2 + 5

x3

7 + 4

x2 + 3 x3

= lim

x→+∞ 2 +

lim

x→+∞

5 x3 lim

x→+∞ 7 +

lim

x→+∞

4 x2 + lim

x→+∞

3 x3 = 2 + 0 7 + 0 + 0 = 2 7. Hence lim

x→+∞

2x3 + 5 7x3 + 4x + 3 = 2 7.

D.L. White (Kent State University) 6 / 13

Examples

Example

Find lim

x→−∞

3x4 +

3

√x 5 − 8x4 if the limit exists.

D.L. White (Kent State University) 7 / 13

Examples

Solution

Observe that

3

√x = x1/3 is defined for x < 0 and the highest power of x in the denominator is x4. We have lim

x→−∞

3x4 +

3

√x 5 − 8x4 = lim

x→−∞

(3x4 +

3

√x) 1

x4

(5 − 8x4) 1

x4

= lim

x→−∞ 3x4 x4 + x1/3 x4 5 x4 − 8x4 x4

= lim

x→−∞

3 +

1 x11/3 5 x4 − 8

= 3 + 0 0 − 8 = −3 8. Hence lim

x→−∞ 3x4+ 3 √x 5−8x4

= −3

8.

D.L. White (Kent State University) 8 / 13

Examples

Example

Find lim

x→+∞

5x4 + x3 2x5 − 4x2 if the limit exists.

D.L. White (Kent State University) 9 / 13

Examples

Solution

The highest power of x in the denominator is x5. We have lim

x→+∞

5x4 + x3 2x5 − 4x2 = lim

x→+∞

(5x4 + x3) 1

x5

(2x5 − 4x2) 1

x5

= lim

x→+∞ 5x4 x5 + x3 x5 2x5 x5 − 4x2 x5

= lim

x→+∞ 5 x + 1 x2

2 − 4

x3

= 0 + 0 2 − 0 = 0 2 = 0. Hence lim

x→+∞ 5x4+x3 2x5−4x2 = 0.

D.L. White (Kent State University) 10 / 13

Examples

Example

Find lim

x→−∞

3x5 + 2 4x2 − 7x if the limit exists.

D.L. White (Kent State University) 11 / 13

Examples

Solution

The highest power of x in the denominator is x2. We have lim

x→−∞

3x5 + 2 4x2 − 7x = lim

x→−∞

(3x5 + 2) 1

x2

(4x2 − 7x) 1

x2

= lim

x→−∞ 3x5 x2 + 2 x2 4x2 x2 − 7x x2

= lim

x→−∞

3x3 + 2

x2

4 − 7

x

Now as x → −∞,

2 x2 → 0 and 4 − 7 x → 4 − 0 = 4, while x3 → −∞.

Hence we have an infinite limit at infinity; lim

x→−∞ 3x5+2 4x2−7x = −∞.

D.L. White (Kent State University) 12 / 13

General Conclusion

We can generalize these methods to obtain the following result.

Theorem

Let f (x) = P(x)

Q(x) be a rational function (P(x), Q(x) are polynomials).

If the degrees of P(x) and Q(x) are equal, then lim

x→+∞ f (x) =

lim

x→−∞ f (x) = p

q , where p is the coefficient of the highest degree term of P(x) and q is the coefficient of the highest degree term of Q(x). If the degree of P(x) is less than the degree of Q(x), then lim

x→+∞ f (x) = 0 and

lim

x→−∞ f (x) = 0.

If the degree of P(x) is greater than the degree of Q(x), then lim

x→+∞f (x) and

lim

x→−∞f (x) are infinite.

D.L. White (Kent State University) 13 / 13