Chapter 2.1, Problem 67E

To determine

To find: The value of $\underset{x\to 0}{\lim }\left(x^2\sin \left(\frac{1}{x^2}\right)\right)$ with the help of graph and check the results with the help of sandwich theorem.

Answer to Problem 67E

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

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\left(x^2\sin \left(\frac{1}{x^2}\right)\right)$ .

Calculation:

To find the limit of the function graphically, let $y=x^2\sin \left(\frac{1}{x^2}\right)$ .

To graph a function $y=x^2\sin \left(\frac{1}{x^2}\right)$ , follow the steps using graphing calculator.

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

The display will show the equation,

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

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)

The function $x^2\sin \left(\frac{1}{x^2}\right)$ tends to $0$ as $x$ tends to $0$ from both the sides. So, the value of $\underset{x\to 0}{\lim }\left(x^2\sin \left(\frac{1}{x^2}\right)\right)$ graphically is $0$ .

Check the limit with the help of sandwich theorem.

As the value of function $\sin x$ lies between $-1$ and $1$ . So, the value of function $x^2\sin \left(\frac{1}{x^2}\right)$ lies between $-x^2$ and $x^2$ .

  $-x^2\le x^2\sin \left(\frac{1}{x^2}\right)\le x^2$

Calculate the limit of $-x^2$ and $x^2$ .

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

And,

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

The limit $\underset{x\to 0}{\lim }\left(x^2\sin \left(\frac{1}{x^2}\right)\right)$ lies between $0$ and $0$ . So, the limit $\underset{x\to 0}{\lim }\left(x^2\sin \left(\frac{1}{x^2}\right)\right)$ is equal to $0$ .

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