Chapter 11, Problem 6RE

Solution

To calculate: The limit of $f\left(x\right)=\frac{\sin 5x}{x}$ at $x=0$ .

The limit of the given function at $x=0$ is 5.

Given information: The given function is $f\left(x\right)=\frac{\sin 5x}{x}$ .

Formula;

The limit $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ .

Multiply and divide the function by 5 and simplify the obtained limit using the formula $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ to obtain the limit of the given function at $x=0$ .

  $\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{5\sin 5x}{5x}\\ =5\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\\ =5\cdot 1\\ =5\end{array}$

Thus, the limit of the given function at $x=0$ is 5.

Draw the graph of the function $f\left(x\right)=\frac{\sin 5x}{x}$ to verify the obtained limit.

  

Figure (1)

From the graph, it can be observed that the function approaches 5 at $x=0$ . Hence, the required limit is 5.