Question

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

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

Answer 1

The correct option is C $e^2$

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