Question

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

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

Answer 1

The correct option is C

$e^2$

${lim}_{x\to 0}{\left(\frac{f(1+x)}{f\left(1\right)}\right)}^{\frac{1}{x}}$

$=e^{{lim}_{x\to 0}\frac{1}{x}\left[\ln \right(f(1+x))-\ln (f\left(1\right)\left)\right]}$

$=e^{{lim}_{x\to 0}\frac{1}{f(1+x)}\times f^{\prime }(1+x)}$ [Applying L'Hospital Rule]

$=e^{\frac{f^{\prime }\left(1\right)}{f\left(1\right)}}$

$=e^{\frac{6}{3}}=e^2$