The value of $f\left(0\right)$ so that $f\left(x\right)={\left(1+3x\right)}^{\frac{1}{x}}$ is continuous at $x=0$ is

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

           =$\underset{x\to 0}{\lim }{(1+3x)}^{\frac{1}{x}}= e^{\underset{x\to 0}{\lim }\frac{1}{x}(1+3x-1)}$                            

$\Rightarrow f\left(0\right)=e^3$