   Chapter 2.3, Problem 37E

Chapter
Section
Textbook Problem

If 4x – 9 ≤ f(x)≤ x2 – 4x + 7 for x ≥ 0, find lim x → 4 f ( x )

To determine

To evaluate: The limit value of the function f(x) as x approaches 4.

Explanation

Given:

The inequality, 4x9f(x)x24x+7 for x0.

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

Calculation:

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

The given inequality becomes, 4x9f(x)x24x+7.

When the limit x approaches to 4, the inequality becomes,

limx44x9limx4f(x)limx4x24x+7

Let g(x)=4x9 and h(x)=x24x+7

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