Question

Let $f:R\to R$ be such that f(1)=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

Answer: (A), (C)

The correct options are
A 1
C $e^2$

$\underset{x\to 0}{lim}{\left(\frac{f(1+x)}{f\left(1\right)}\right)}^{1/x}$, $\left[1^{\infty }\right]$ form

$=\underset{x\to 0}{lim}{(1+u)}^{\frac{1}{u}\frac{f(1+x)-f\left(1\right)}{1+x-1}\times \frac{1}{f\left(1\right)}}$

Where $u=\frac{f(1+x)-f\left(1\right)}{f\left(1\right)}=\frac{f(1+x)-f\left(1\right)}{1+x-1}\times \frac{x}{3f\left(1\right)}\to \frac{1}{3f\left(1\right)}{f}^{\prime }\left(1\right)\times 0=0$ as $x\to 0$

$=\underset{u\to 0}{lim}{(1+u)}^{\frac{1}{u}\frac{f(1+x)-f\left(1\right)}{1+x-1}\times \frac{1}{f\left(1\right)}}$

$=e^{f^{\prime }\left(1\right)/f\left(1\right)}=e^{6/3}=e^2$
Note that we cannot take the logarithm as the function may take negative values and also the L Hospital's rule is not applicable.