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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Prove that lim x x e sin ( π / x ) = 0.

To determine

To prove: The value of limx0+xesin(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.”

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist, then

Limit law 3: limxa[cf(x)]=climxaf(x)

Limit law 10: limxaxn=an where n is a positive integer, if n is even, assume that a>0.

Proof:

It is trivial that, the value of limx0+xesin(πx) does not exist.

limx0+xesin(πx)=limx0+xlimx0+(esin(πx))=0esin(π0)

Thus, the limit of the function does not exist.

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

Since the sine function lies between 1 and 1,consider 1sin(πx)1.

Take the exponential of the inequality, e1esin(πx)e1

Any inequality remains true when multiplied by a positive number. Since x0 for all x, multiply each side of the inequalities by x

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