   Chapter 2, Problem 1P ### 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

# Evaluate lim x → 1 x 3 − 1 x − 1

To determine

To evaluate: The value of limx1x31x1

Explanation

Result used:

Difference of squares formula: (a2b2)=(a+b)(ab)

Difference of cubes formula: (a3b3)=(ab)(a2+ab+b2)

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limxaf(x)=f(a).

Calculation:

Obtain the limit of the function f(x)=x31x1 as x approaches 1 as follows.

Consider the function f(x)=x31x1.

Take the limit of the function as x approaches 1.

limx1f(x)=limx1x31x1=limx1(x13)1(x12)1=limx1(x26)1(x36)1=limx1(x16)21(x16)31

Let x6=t. This implies that, x=t6. Then t approaches 1 as x approaches 1

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