Let $f:R\to R$ be such that $f\left(1\right)=3 \text{and }{f}^{\mathrm{\prime }}\left(1\right)=6\text{,}$ then  $\underset{x\to 0}{lim}{\left(\frac{f(1+x)}{f\left(1\right)}\right)}^{1/x}$ is equal to

$$\begin{gathered} \underset{x\to 0}{lim}{\left[\frac{f(1+x)}{f\left(1\right)}\right]}^{1/x}=e^{\underset{x\to 0}{\lim }\frac{\left[\log \left(f\right(1+x\left)\right)-\log \left(f\right(1\left)\right)\right]}{x}} \\ =e^{\underset{x\to 0}{\lim }\left[\frac{1}{f(1+x)}f\text{'}(1+x)-0\right]} \\ =e^{\frac{f\text{'}\left(1\right)}{f\left(1\right)}} \\ =e^2 \end{gathered}$$