Chapter 2.1, Problem 31E

To determine

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

Answer to Problem 31E

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

Explanation of Solution

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

Calculation:

Let $y=\frac{\sin x}{2x^2-x}$ to find the value of $\underset{x\to 0}{\lim }\left(\frac{\sin x}{2x^2-x}\right)$ graphically.

To graph a function $y=\frac{\sin x}{2x^2-x}$ , 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 $y=\frac{\sin x}{2x^2-x}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{SIN}}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{2}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{\text{X}}\boxed{)}$ .

The display will show the equation,

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

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

Check the limit algebraically. Use the property of limit.

  $\underset{x\to 0}{\lim }\left(\frac{\sin \left(ax\right)}{ax}\right)=1$

Simplify the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{\sin x}{2x^2-x}\right)=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x\left(2x-1\right)}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)\times \underset{x\to 0}{\lim }\left(\frac{1}{2x-1}\right)\\ =1\times \frac{1}{2\times 0-1}\\ =-1\end{array}$

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

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