Question

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

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

Answer 1

The correct option is C $e^2$

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

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

Hence, option C is correct.