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 5P

(a)

To determine

To evaluate: The limit of the function limx0xx.

Expert Solution

Answer to Problem 5P

The limit of the function does not exist.

Explanation of Solution

Definition used:

Greatest integer function:x= the largest integer that is less than or equal to x.

Calculation:

Obtain the limit of the function limx0xx.

Let the function f(x)=xx.

For 0<x<1,

By the definition of greatest integer function, x=0.

Thus, xx=0. (1)

For 1<x<0,

By the definition of greatest integer function, x=1.

Thus, xx=1x. (2)

The left hand limit of the function as x approaches 0 is computed as follows,

limx0f(x)=limx0xx=limx0(1x)[by equation (2)]=10=

The right hand limit of the function as x approaches 0 is computed as follows,

limx0+f(x)=limx0+xx=limx0+(0)[by equation (1)]=0

Since the left and right hand limits are not equal.

Thus, the limit of the function limx0xx does not exist.

(b)

To determine

To evaluate: The limit of the function limx0x1x.

Expert Solution

Answer to Problem 5P

The limit of the function is 1.

Explanation of Solution

Result used: Squeeze theorem

“Suppose that g(x)f(x)h(x) for all x in some open interval containing c except possibly at c itself, if limxcg(x)=L=limxch(x) then limxcf(x)=L”.

Calculation:

Obtain the limit of the function limx0x1x.

Let the function f(x)=x1x.

For x>0 such that,

1x11x1xx(1x1)x1xx(1x)x(1xx)x1xx(1x)1xx1x1

Take the right hand limit of the above inequality as x approaches 0,

limx0+(1x)limx0+x1xlimx0+1

Here, limx0+(1x)=1 and limx0+(1)=1.

Therefore, by the Squeeze theorem, limx0+x1x=1.

For x<0 such that

1x11x1xx(1x1)x1xx(1x)x(1xx)x1xx(1x)1xx1x1

Take the left hand limit of the above inequality as x approaches 0,

limx0(1x)limx0x1xlimx01

Here, limx0(1x)=1 and limx0(1)=1.

Therefore, by the Squeeze theorem, limx0x1x=1.

Since the left and right hand limits are equal, the limit exists and equal to 1. That is, limx0x1x=1.

Thus, the limit of the function is 1.

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