A function $f:R\to R$ is such that $f\left(1\right)=3$ and $f^{\prime }\left(1\right)=6$. Then $\underset{x\to 0}{lim}{\left[\frac{f(1+x)}{f\left(1\right)}\right]}^{1/x}=$

The correct option is $\left(B\right)$

Given : $f\left(1\right)=3,f^{\prime }\left(1\right)=6$.....(1)

Let say

$y={lim}_{x\to 0}{\left(\frac{f(1+x)}{f\left(1\right)}\right)}^{\frac{1}{x}}$

taking $loge$ both side -

$lny={lim}_{x\to 0}ln{\left(\frac{f(1+x)}{f\left(1\right)}\right)}^{1/x}$

$={lim}_{x\to 0}\frac{1}{x}ln\left(\frac{f(1+x)}{f\left(1\right)}\right)$ $[\because \ln m^n=nlnm]$

$lny={lim}_{x\to 0}\left[\frac{ln\left(f\right(1+x\left)\right)-lnf\left(1\right)}{x}\right]$ $[\because \ln \frac{m}{n}=\ln m-\ln n]$

Now apply L.Hospital rule-

$lny={lim}_{x\to 0}\frac{\left[\frac{1}{f(1+x)}{f}^{\prime }\right(1+x)-0]}{1}=\frac{1f^{\prime }\left(1\right)}{f\left(1\right)}$

$lny=\frac{6}{3}=2$ [From (1)]

$\Rightarrow y=e^2$

Hence the correct answer is option 'B'