   Chapter 2.3, Problem 35E

Chapter
Section
Textbook Problem

Use the Squeeze Theorem to show that lim x → 0 ( x 2 cos 20 π x ) = 0 . Illustrate by graphing the functions f(x) = –x2, g(x) = x2cos 20πx, and h(x) = x2 on the same screen.

To determine

To show: The limit of the function limx0x2cos20πx=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:

Apply the Squeeze Theorem and obtain a function f smaller than g(x)=x2cos20πx and a function h bigger than g(x)=x2cos20πx such that both f(x) and h(x) approaches 0.

Since the cosine function is lies between 1 and 1, 1x2cos20πx1.

Any inequality remains true when multiplied by a positive number. Since x20 for all x, multiply each side of the inequalities by x4.

1×x2x2×cos20πx1×x2x2x2cos20πxx2

Let f(x)=x2, g(x)=x2cos20πx and h(x)=x2

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