If _{$f\left(x\right)={\left(1+3\tan \left(x^2\right)\right)}^{\frac{1}{6x^2}}$}  is continuous at x = 0 then f(0) =

   $f\left(0\right)=\underset{x\to 0}{Lt}f\left(x\right)=\underset{x\to 0}{Lt}{\left(1+3\tan \left(x^2\right)\right)}^{\frac{1}{6x^2}}$                      $\underset{x\to c}{\lim }f{\left(x\right)}^{g\left(x\right)} \to 1^{\infty } then \underset{x\to c}{\lim }f{\left(x\right)}^{g\left(x\right)}=e^{\underset{x\to c}{\lim }g\left(x\right)\left\{f\right(x)-1\}}$

         $=e^{\underset{x\to 0}{Lt}\frac{1}{6x^2}(1+3\tan \left(x^2\right)-1)}=e^{\underset{x\to 0}{\lim }\frac{3 {\tan }^{2}x}{6x^2}}$

        $=e^{\frac{1}{2}}=\sqrt{e}$