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

Solution :
Given that $$f:R \to R$$   such that
$$\eqalign{ & f\left( 1 \right) = 3\,\,{\text{and}}\,\,f'\left( 1 \right) = 6 \cr & {\text{Then}}\,\,\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{f\left( {1 + x} \right)}}{{f\left( 1 \right)}}} \right)^{\frac{1}{x}}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \,\frac{1}{x}\left[ {\log \,f\left( {1 + x} \right) - \log \,f\left( 1 \right)} \right]}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{f\left( {1 + x} \right)}}f'\left( {1 + x} \right)}}{1}}}\,\,\,\,\left[ {{\text{Using L'Hospital rule}}} \right] \cr & = {e^{\frac{{f'\left( 1 \right)}}{{f\left( 1 \right)}}}} \cr & = {e^{\frac{6}{3}}} \cr & = {e^2} \cr} $$