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

$\underset{x\to 0}{\mathrm{Lt}}f\left(x\right)=f\left(0\right)=12$

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

$\begin{array}{c}\underset{x\to 0}{lim} \frac{{\left({\displaystyle \frac{e^x-1}{x^2}}\right)}^2}{{\displaystyle \frac{\sin \left(\frac{x}{k}\right)}{x}\frac{ \log \left(1+\frac{x}{4}\right)}{x}}}\end{array}$

$\frac{1}{{\displaystyle \frac{1}{k}\frac{1}{4}}}=4k$

On applying a limit we get $4k=12\Rightarrow k=3$