Let k be a non-zero real number. If $f\left(x\right)=\left\{\begin{array}{l}\frac{{\left(e^x-1\right)}^2}{\sin \left(\frac{x}{k}\right). \log \left(1+\frac{x}{4}\right)},x\ne 0\\ 12,x=0\end{array}\right.$is a continuous function then the value of k is

   

 $\begin{array}{l}\underset{x\to 0}{lim} f\left(x\right)=f\left(0\right)=12\\ \Rightarrow \underset{x\to 0}{lim} f\left(x\right)=\underset{x\to 0}{lim} \frac{{\left(e^x-1\right)}^2}{\sin \left(\frac{x}{k}\right)\log \left(1+\frac{x}{4}\right)}\\ \Rightarrow \underset{x\to 0}{lim} f\left(x\right)=\underset{x\to 0}{lim} \frac{{\left(\frac{e^x-1}{x}\right)}^{2}4k}{\sin \left(\frac{x}{k}\right)\cdot \frac{4}{x}\log \left(1+\frac{x}{4}\right)}=4k\\ \end{array}$

on applying limit we get