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 22RE
To determine

To prove: The limit of the function limx0x2cos(1x2)=0.

Expert Solution

Explanation of Solution

Theorem used: The Squeeze Theorem

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

Proof:

It is trivial that, the value of limx0cos(1x2) does not exist.

limx0x2cos(1x2)=0=limx0x2limx0cos(1x2)=(0)2cos(10)

Thus, the limit of the function does not exist.

Apply the Squeeze Theorem and obtain a function f smaller than g(x)=x2cos(1x2) and a function h bigger than g(x)=x2cos(1x2) such that both f(x) and h(x) approaches 0.

Since the cosine function is lies between 1 and 1, 1cos(1x2)1.

Any inequality remains true when multiplied by a positive number. Since x20 for all x, multiply each side of the inequalities by x2.

1×x2x2×cos(1x2)1×x2x2x2cos(1x2)x2

When the limit x approaches zero, the inequality becomes,

limx0(x2)limx0(x2cos(1x2))limx0(x2)

Graph:

Use the online graphing calculator to draw the graph of the function as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 2, Problem 22RE

From Figure 1, it is observed that limx0(x2)=0 and limx0(x2)=0.

Let f(x)=x2, g(x)=x2cos(1x2) and h(x)=x2.

If f(x)g(x)h(x) when x approaches 0 and limx0(x2)=limx0(x2)=0, then by Squeeze Theorem the limit of the function g(x) is zero.

That is, limx0x2cos(1x2)=0.

Hence the required proof is obtained.

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