BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 2.3, Problem 28E
To determine

To show: The limit of the function limx0(x3+x2sin(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.”

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.

1×x3+x2x3+x2×sin(πx)1×x3+x2x3+x2x3+x2sin(πx)x3+x2

When the limit x approaches to zero, the inequality becomes,

limx0(x3+x2)limx0x3+x2sin(πx)limx0x3+x2

Graph:

Sketch the graph of the function by using the online graphing calculator as shown below in Figure (1).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 2.3, Problem 28E

From the graph, it is observed that limx0(x3+x2)=0 and limx0(x3+x2)=0.

If f(x)g(x)h(x) when x approaches 0 and limx0(x3+x2)=limx0(x3+x2)=0, then by Squeeze Theorem the limit of the function g(x) is zero.

That is, limx0x3+x2sin(πx)=0.

Hence showed.

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