Chapter 2.3, Problem 27E

To determine

To show: The limit of the function $\underset{x\to 0}{\lim }{x}^2\cos 20\pi x=0$.

Explanation of Solution

Theorem used: The Squeeze Theorem

“If $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ when x is near a (except possibly at a) and $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }h\left(x\right)=L$ then $\underset{x\to a}{\lim }g\left(x\right)=L$.”

Proof:

Apply the Squeeze Theorem and obtain a function f smaller than $g\left(x\right)=x^2\cos 20\pi x$ and a function h bigger than $g\left(x\right)=x^2\cos 20\pi x$ such that both $f\left(x\right)$ and $h\left(x\right)$ approaches 0.

Since the cosine function is lies between $-1$ and 1, $-1\le x^2\cos 20\pi x\le 1$.

Any inequality remains true when multiplied by a positive number. Since $x^2\ge 0$ for all x, multiply each side of the inequalities by $x^4$.

$\begin{array}{l}-1\times x^2\le x^2\times \cos 20\pi x\le 1\times x^2\\ -x^2\le x^2\cos 20\pi x\le x^2\end{array}$

Let $f\left(x\right)=-x^2$, $g\left(x\right)=x^2\cos 20\pi x$ and $h\left(x\right)=x^2$.

When the limit x approaches zero, the inequality becomes,

$\underset{x\to 0}{\lim }\left(-x^2\right)\le \underset{x\to 0}{\lim }\left(x^2\cos 20\pi x\right)\le \underset{x\to 0}{\lim }\left(x^2\right)$

Graph:

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

From the graph, it is observed that $\underset{x\to 0}{\lim }\left(-x^2\right)=0$ and $\underset{x\to 0}{\lim }\left(x^2\right)=0$.

If $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ when x approaches 0 and $\underset{x\to 0}{\lim }\left(-x^2\right)=\underset{x\to 0}{\lim }\left(x^2\right)=0$, then by Squeeze Theorem the limit of the function $g\left(x\right)$ is zero.

That is, $\underset{x\to 0}{\lim }{x}^2\cos 20\pi x=0$.

Hence showed.