# Let $f:R\rightarrow R$ be such that $f(1)=3$ and $f'(1)=6$. Then $\lim\limits_{x\to 0}\big[\large\frac{f(1+x)}{f(1)}\big]^{\Large\frac{1}{x}}$ equals

$(a)\;1\qquad(b)\;e^{\Large\frac{1}{2}}\qquad(c)\;e^2\qquad(d)\;e^3$

## 1 Answer

Given that $f: R\rightarrow R$ such that
$f(1)=3$ and $f'(1)=6$
Then $\lim\limits_{x\to 0}\bigg[\large\frac{f(1+x)}{f(1)}\bigg]^{1/x}$
$\Rightarrow e^{\lim\limits_{x\to 0}\large\frac{1}{x}[\log f(1+x)-\log f(1)]}$
$\Rightarrow e^{\large\lim\limits_{x\to 0}\Large\frac{1/f(1+x)f'(1+x)}{1}}$
Using L Hospital rule
$\Rightarrow e^{\large\frac{f'(1)}{f(1)}}$
$\Rightarrow e^{\Large\frac{6}{3}}$
$\Rightarrow e^2$
Hence (c) is the correct answer.
answered Dec 31, 2013

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