Question

Let $f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x}, & x \neq 0 \\ k, & x=0\end{array}\right.$. If $f(x)$ is continuous at $x=0$, then $k$ is equal to

(a) $\frac{\pi}{5}$

(b) $\frac{5}{\pi}$

(c) 1

(d) 0

Answer 1

SOLUTION — Since, $f(x)$ is continuous at $x=0$, therefore

$\lim _{x \rightarrow 0} f(x) =f(0) \\$

$\Rightarrow  \lim _{x \rightarrow 0} \frac{\sin \pi x}{5 x}  =k \\$

$\Rightarrow \quad  \lim _{x \rightarrow 0}\left(\frac{\sin \pi x}{\pi x}\right) \frac{\pi}{5}  =k \\$

$\Rightarrow \quad  \text { (1) } \frac{\pi}{5}=k \Rightarrow k  =\frac{\pi}{5} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)$

So, The correct option of this question will be (A).