Chapter 2.2, Problem 9E

To determine

To find: The value of $\underset{x\to \infty }{\lim }\left(\frac{1-\cos x}{x^2}\right)$ and confirm the answer with the help of Sandwich theorem.

Answer to Problem 9E

The value of $\underset{x\to \infty }{\lim }\left(\frac{1-\cos x}{x^2}\right)$ is $0$ .

Explanation of Solution

Given information:The limit is $\underset{x\to \infty }{\lim }\left(\frac{1-\cos x}{x^2}\right)$ .

Calculation:

To graph a function $y=\frac{1-\cos x}{x^2}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{1-\cos x}{x^2}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{1}}\boxed{-}\boxed{\text{COS}}\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(1-\cos \left(\text{X}\right)\right)/\left({\text{X}}^2\right)$

Press the window key and adjust the window to $\left[-5,5\right]$ by $\left[-2,2\right]$ .

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from graph, the function $f\left(x\right)=\frac{1-\cos x}{x^2}$ tends to $0$ as $x$ tends to $\infty$ .So, the value of $\underset{x\to \infty }{\lim }\left(\frac{1-\cos x}{x^2}\right)$ is $0$ .

Check the answer by sandwich theorem.

It is known that the function cosine lies between $-1$ and $1$ . So,

  $\begin{array}{c}-1\le \cos x\le 1\\ -1\le -\cos x\le 1\\ 1-1\le 1-\cos x\le 1+1\\ 0\le 1-\cos x\le 2\end{array}$

Divide all the sides of inequality by $x^2$ .

  $\begin{array}{l}0\le 1-\cos x\le 2\\ 0\le \frac{1-\cos x}{x^2}\le \frac{2}{x^2}\end{array}$

Now calculate the limits.

  $\underset{x\to \infty }{\lim }0=0$ and $\underset{x\to \infty }{\lim }\left(\frac{2}{x^2}\right)=0$

Thus,

  $\begin{array}{c}\underset{x\to \infty }{\lim }0\le \underset{x\to \infty }{\lim }\frac{1-\cos x}{x^2}\le \underset{x\to \infty }{\lim }\frac{2}{x^2}\\ 0\le \underset{x\to \infty }{\lim }\frac{1-\cos x}{x^2}\le 0\end{array}$

Therefore,the value of $\underset{x\to \infty }{\lim }\left(\frac{1-\cos x}{x^2}\right)$ is $0$ .