Question

A function $f$ : $R\to R$ is such that $f\left(1\right)=3$ and $f^{\prime }\left(1\right)=6$. Then ${lim}_{x\to 0}{(\frac{f(1+x)}{f\left(1\right)})}^{\frac{1}{x}}=$

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

Answer 1

The correct option is B $e^2$

$\underset{x\to 0}{lim}f(x{)}^{g\left(x\right)}=e^{(\underset{x\to 0}{lim}\left(f\right(x)-1).g\left(x\right))}$

Using above formula we get,

$e^{{lim}_{x\to 0}\frac{f(1+x)-f\left(1\right)}{xf\left(1\right)}}$
$=e^{\frac{f^{\prime }\left(1\right)}{f\left(1\right)}}$
$=e^{\frac{6}{3}}=e^2$