Question

Let $f: R \rightarrow R$ be such that $f(1)=3$ and $f'(1)=6 .$ Then, $\displaystyle\lim _{x \rightarrow 0}\left[\frac{f(1+x)}{f(1)}\right]^{1 / x}$ equals

• A. $ 1 $
• B. $e^{1/2}$
• C. $e^{2}$
• D. $e^{3}$

Answer 1

Answer: (C)

Let $y=\left[\frac{f(1+x)}{f(1)}\right]^{1 / x}$
$\Rightarrow \log y=\frac{1}{x}[\log f(1+x)-\log f(1)]$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \log y=\displaystyle\lim _{x \rightarrow 0} \frac{[\log f(1+x)-\log f(1)]}{x}$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \log y=\displaystyle\lim _{x \rightarrow 0}\left[\frac{1}{f(1+x)} f'(1+x)\right]$
(using L' Hospital's rule)
$=\frac{f'(1)}{f(1)}=\frac{6}{3}=2$
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} y=e^{2}$