Question

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

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

Answer 1

Answer: (C)

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