Let $f:R\to R$ be such that f(1) = 3 and f' (1) = 6.Then $\underset{x\to 0}{lim} {\left\{\frac{f(1+x)}{f\left(1\right)}\right\}}^{1/x}$ equals. 

Solution:

We have,

$\begin{array}{l}\underset{x\to 0}{lim} {\left\{\frac{f(1+x)}{f\left(1\right)}\right\}}^{1/x}\\ =\underset{x\to 0}{lim} {\left\{1+\frac{f(1+x)-f\left(1\right)}{f\left(1\right)}\right\}}^{1/x}\\ =e\underset{x\to 0}{lim} \frac{f(1+x)-f\left(1\right)}{xf\left(1\right)}\\ =e^{\frac{1}{f\left(1\right)}\underset{x\to 0}{lim} \frac{f(x+i)-f\left(1\right)}{x}}\\ =e^{\frac{f^{\mathrm{\prime }}\left(1\right)}{f\left(1\right)}}=e^{6/3}=e^2\left[\because \underset{x\to 0}{lim} \frac{f(1+x)-f\left(1\right)}{x}=f^{\mathrm{\prime }}\left(1\right)\right]\end{array}$