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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Use the Squeeze Theorem to show that

lim x 0 x 3 + x 2 sin π x = 0

Illustrate by graphing the functions f, q, and h (in the notation of the Squeeze Theorem) on the same screen.

To determine

To show: The limit of the function limx0(x3+x2sin(πx))=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)=x3+x2sin(πx) and a function h bigger than g(x)=x3+x2sin(πx) such that both f(x) and h(x) approaches 0.

Since the cosine function is lies between 1 and 1, 1sin(πx)1.

Let f(x)=x3+x2, g(x)=x3+x2sin(πx) and h(x)=x3+x2

Any inequality remains true when multiplied by a positive number. Since x3+x20 for all x, multiply each side of the inequalities by x3+x2

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