   Chapter 2.3, Problem 39E ### 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

# Prove that lim x → 0 x 4 cos 2 x = 0.

To determine

To prove: The limit of the function limx0x4cos2x=0.

Explanation

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 limx0cos2x does not exist.

limx0x4cos2x=limx0x4limx0cos(2x)=(0)4cos(20)

Thus, the limit of the function does not exist.

Apply the Squeeze Theorem and obtain a function f smaller than g(x)=x4cos(2x) and a function h bigger than g(x)=x4cos(2x) such that both f(x) and h(x) approaches 0.

Since the cosine function is lies between 1 and 1, 1cos(2x)1.

Any inequality remains true when multiplied by a positive number. Since x40 for all x, multiply each side of the inequalities by x4

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