Question:
If $f(x)=\left\{\begin{array}{cl}\frac{x}{\sin 3 x}, & x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at $x=0$, then write the value of $k$.
Solution:
If $f(x)$ is continuous at $x=0$, then
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{\sin 3 x}=k$
$\Rightarrow \lim _{x \rightarrow 0} \frac{1}{\frac{\sin 3 x}{x}}=k$
$\Rightarrow \lim _{x \rightarrow 0} \frac{1}{\frac{3 \sin 3 x}{3 x}}=k$
$\Rightarrow \frac{1}{3}\left(\frac{1}{\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}\right)=k$
$\Rightarrow k=\frac{1}{3}$