Chapter 2.3, Problem 27E

To determine

To find: The formula for the extended function $f\left(x\right)=\frac{\sin x}{x}$ such that $f\left(x\right)$ is continuous at $x=0$ .

Answer to Problem 27E

The formula for the extended function $f\left(x\right)=\frac{\sin x}{x}$ such that $f\left(x\right)$ is continuous at $x=0$ is $y=\{\begin{array}{l}\frac{\sin x}{x},x\ne 0\\ 1,x=0\end{array}$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{\sin x}{x}$ and the point is $x=0$ .

Calculation:

The value of the function $f\left(x\right)=\frac{\sin x}{x}$ at $x=0$ doesn’t exist.

As the function $f\left(x\right)=\frac{\sin x}{x}$ has removable continuity at $x=0$ as the limit value of $f\left(x\right)=\frac{\sin x}{x}$ at $x=0$ is equal to $1$ and the value of $f\left(0\right)$ is undefined.

To make the function $f\left(x\right)=\frac{\sin x}{x}$ continuous at $x=0$ set the value of $f\left(0\right)$ equal to $1$ .

So, the extended function $y=\{\begin{array}{l}\frac{\sin x}{x},x\ne 0\\ 1,x=0\end{array}$ is continuous at $x=0$ .

Therefore, the formula for the extended function $f\left(x\right)=\frac{\sin x}{x}$ such that $f\left(x\right)$ is continuous at $x=0$ is $y=\{\begin{array}{l}\frac{\sin x}{x},x\ne 0\\ 1,x=0\end{array}$ .