BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 2, Problem 1P
To determine

To evaluate: The value of limx1x31x1

Expert Solution

Answer to Problem 1P

The limit of the function is 23_

Explanation of Solution

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.

On substitution,

limx1x31x1=limt1((t6)16)21((t6)16)31=limt1(t)21(t)31

Apply the difference of squares and cubes formulae in the numerator and the denominator as follows,

limx1x31x1=limt1(t+1)(t1)(t1)(t2+t+1)=limt1(t+1)(t2+t+1)=((1)+1)((1)2+(1)+1)[by direct substitution]=23

Thus, the limit of the function is 23_.

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