Chapter 13.1, Problem 9E

To determine

To estimate: The value of the limit using values of $f(x)$ for $x=\pm 1,\pm 0\cdot 5,\pm 0\cdot 1,\pm 0\cdot 005,\pm 0\cdot 01$

Answer to Problem 9E

The value of the limit is $1$ .

Explanation of Solution

Given information:

Limit is $\underset{x\to 0}{{\displaystyle \lim }}\frac{\sin x}{x}$

Calculation:

Limit is $\underset{x\to 0}{{\displaystyle \lim }}\frac{\sin x}{x}$

1. When $x=1$

  $1$ is in degree we have to change it to radian

  $\begin{array}{l}x=1\times \frac{\pi }{180}=0\cdot 01745\\ \frac{\sin x}{x}=\frac{\sin \left(0\cdot 01745\right)}{0\cdot 01745}\Rightarrow f(x)=0\cdot 99994\end{array}$

2. When $x=-1$

  $-1$ is in degree we have to change it to radian

  $\begin{array}{l}x=-1\times \frac{\pi }{180}=-0\cdot 01745\\ \frac{\sin x}{x}=\frac{\sin \left(-0\cdot 01745\right)}{-0\cdot 01745}=\frac{-\sin \left(-0\cdot 01745\right)}{-0\cdot 01745}=0\cdot 99994\end{array}$

So, $f(x)=f(-x)$

3. When $x=0\cdot 5$

  $\begin{array}{l}x=\pm 0\cdot 5\times \frac{\pi }{180}=\pm 0\cdot 00872\\ f(x)=\frac{\sin x}{x}=0\cdot 99998\end{array}$

4. $x=\pm 0\cdot 1\Rightarrow$ In degree $x=\pm 0\cdot 001745$

  $f(x)=\frac{\sin x}{x}=0\cdot 99999$

5. $x=\pm 0\cdot 05\Rightarrow$ In degree $x=\pm 0\cdot 00087$

  $f(x)=\frac{\sin x}{x}=0\cdot 99999$

6. $x=\pm 0\cdot 01\Rightarrow$ In degree $x=\pm 0\cdot 00017$

  $f(x)=\frac{\sin x}{x}=0\cdot 99999$

  $\begin{array}{l}x=\pm 1\Rightarrow f(x)=0\cdot 99994\\ x=\pm 0\cdot 5\Rightarrow f(x)=0\cdot 99998\\ x=\pm 0\cdot 1\Rightarrow f(x)=0\cdot 99999\\ x=\pm 0\cdot 05\Rightarrow f(x)=0\cdot 99999\\ x=\pm 0\cdot 01\Rightarrow f(x)=0\cdot 99999\end{array}$

By observing these values, the estimated value of limit is $1$