Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{\sin \pi x}{5x},x\ne 0\\ k,x=0\end{array}$. If $f\left(x\right)$ is continuous at $x=0$, the value of $k$ is

• A. $\frac{5}{\pi }$
• B. $\frac{\pi }{5}$
• C. Zero
• D. $1$

Answer 1

Answer: (B)

$\frac{\pi }{5}$

For the given function to be continuous,

its $LHL$ and $RHL$ at the given $x=x_o$ and $f\left(x_o\right)$ should be equal.

Now, $f\left(0\right)=k$

Then, we have

$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)\Rightarrow k=\underset{x\to 0}{lim}\frac{\sin \pi x}{5x}$

Applying L'Hospital's rule, we get

$k=\underset{x\to 0}{lim}\frac{\pi \cos \pi x}{5}=\frac{\pi }{5}$

Hence, $k=\frac{\pi }{5}$