If$f\left(x\right)=\begin{array}{c}\left\{\begin{array}{l}\frac{{(5^x-1)}^3}{\sin \left(\frac{x}{a}\right)\log (1+\frac{x^2}{3})}; x\ne 0\\ 9{\left(\log 5\right)}^3 ;x=0\end{array}\right.\\ \end{array}$ is continuous at$x=0$ then value of$\text{'}a\text{'}$ equals

$\underset{x\to 0}{Lt}{\left(\frac{5^x-1}{x}\right)}^3\frac{x^3}{\frac{x\sin \left(\frac{x}{a}\right)}{\left(\frac{x}{a}\right)}}\times \frac{1}{\frac{\log (1+\frac{x^2}{3})}{\frac{x^2}{3}}\times \frac{x^2}{3}}$ 

$=$ $\underset{x\to 0}{Lt}{\left(\frac{5^x-1}{x}\right)}^3\frac{1\times a}{\left(\frac{\sin \left(\frac{x}{a}\right)}{\frac{x}{a}}\right)}.\frac{3}{\frac{\log \left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}}$

$=3a{\left({\log }_{e}5\right)}^3$

Now $f\left(0\right)=9{\left({\log }_{e}5\right)}^3$

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

$\Rightarrow$          $9{\left({\log }_{e}5\right)}^3=3a{\left({\log }_{e}5\right)}^3$

$\Rightarrow$          $a=3$