To prove: The value of lim x → 0 + x e sin ( π x ) = 0 .

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.3, Problem 32E
To determine

To prove: The value of limx→0+xesin(πx)=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.”

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.

e1×xesin(πx)×xe1×xxe1xesin(πx)xe1xexesin(πx)ex

When x approaches to zero, the inequality becomes,

limx0+xelimx0+xesin(πx)limx0+ex

Obtain the value of limx0+xe as follows.

limx0+xe=1elimx0+x (by limit law 3)=1e0 (by limit law 10)=0

Obtain the value of limx0+ex .

limx0ex=elimx0x (by limit law 3)=e0 (by limit law 10)=0

Let f(x)=xe, g(x)=xesin(πx) and h(x)=ex.

Sketch the graph of the functions f(x)=xe, g(x)=xesin(πx) and h(x)=ex by using the online graphing calculator as shown below in Figure (1).

From Figure 1, it is observed that limx0+xe=0 and limx0+ex=0.

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

That is, limx0+xesin(πx)=0.

Hence the required proof is obtained.

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