# The limit value of the function g ( x ) as x approaches 1. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2.3, Problem 30E
To determine

## To evaluate: The limit value of the function g(x) as x approaches 1.

Expert Solution

The limit value of the function g(x) as x approaches 1 is 2.

### Explanation of Solution

Given:

The inequality is, 2xg(x)x4x2+2 for all x.

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist. Then

Limit law 1: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)

Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 3: limxa[cf(x)]=climxaf(x)

Limit law 7: limxac=c

Limit law 8: limxax=a

Limit law 9: limxaxn=an where n is a positive integer.

Theorem used: The Squeeze Theorem

“If f(x)g(x)h(x) when x is near a (except possibly at a) and limxaf(x)=limxah(x)=L then limxag(x)=L.”

Calculation:

Apply the Squeeze Theorem and obtain a function f smaller than g(x) and a function h bigger than g(x) such that both f(x) and h(x) approaches 4.

The given inequality becomes, 2xg(x)x4x2+2.

When the limit x approaches to 1, the inequality becomes,

limx12xlimx1g(x)limx1x4x2+2

Let f(x)=2x and h(x)=x4x2+2.

Obtain the limit value of as x approaches 1 of the function f(x).

limx12x=2limx1x (by limit law 3)=2(1) (by limit law 8)=2

Obtain the limit value of as x approaches 1 of the function h(x).

limx1x4x2+2=limx1(x4x2)+limx12 (by limit law 1)=limx1(x4)limx1x2+limx12 (by limit law 2)=(1)4(1)2+2 (by limit law 9 and 7)=2

If f(x)g(x)h(x) when x approaches 1 and limx12x=limx1(x4x2+2)=2, then by Squeeze Theorem the limit of the function g(x) is 2.

Therefore, limx1g(x)=2.

Thus, the limit value of the function g(x) as x approaches 1 is 2.

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