   Chapter 2, Problem 23RE ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# If 2x – 1 ≤ f(x) ≤ x2 for 0 < x < 3, find lim x → 1 f ( x ) .

To determine

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

Explanation

Given:

The inequality, 2x1f(x)x2 for 0<x<3.

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 1.

The given inequality becomes, 2x1f(x)x2.

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

limx12x1limx1f(x)limx1x2

Let g(x)=2x1 and h(x)=x2

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