Chapter 1, Problem 40RE

To determine

To find: The value of $\left(k\right)$ for which the function $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{2x},x\ne 0\\ k,x=0\end{array}\right.$ is continuous at point $x=0$ .

Answer to Problem 40RE

The value of $k=\frac{1}{2}$ for which the function is continuous at $x=0$ .

Explanation of Solution

Given:

The function is $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{2x},x\ne 0\\ k,x=0\end{array}\right.$ .

Concept used:

The function $g\left(x\right)$ is continuous at some point $x=a$ in domain of $g\left(x\right)$ , if $\underset{x\to a}{\lim }g\left(x\right)=g\left(a\right)$ .

Calculation:

Apply the concept stated above for the function $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{2x},x\ne 0\\ k,x=0\end{array}\right.$ at the point $x=0$ . Therefore,

  $\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)\\ \underset{x\to 0}{\lim }\left(\frac{\sin x}{2x}\right)=k\\ \frac{1}{2}\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=k\\ \frac{1}{2}\left(1\right)=k\mathrm{....}\left(\because \underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1\right)\\ \text{or, }k=\frac{1}{2}\\ k=\frac{1}{2}\end{array}$

Conclusion:

Hence, the value of $k=\frac{1}{2}$ for which the function is continuous at $x=0$ .