• # question_answer Let $f:\text{ }R\to R$ be such that $f\left( 1 \right)=3$ and $f'\left( 1 \right)=6.$ Then $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{f\left( 1+x \right)}{f\left( 1 \right)} \right)}^{1/x}}$ equals A) 1 B) ${{e}^{1/2}}$ C) ${{e}^{2}}$   D) ${{e}^{3}}$

 Given that $f:R\to R$ such that $f(1)=3$and$f'(1)=6$ Then $\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{f\left( 1+x \right)}{f\left( 1 \right)} \right]}^{1/x}}$$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{f(1+x)}f'(1+x)}{1}}}={{e}^{\frac{f'(1)}{f(1)}}}={{e}^{6/3}}={{e}^{2}}$ [Using L Hospital rule]