Question

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

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

Answer 1

The correct option is C

$e^2$

Explanation For The Correct Option:

Evaluating the limit of given function

Given: $\underset{x\to 0}{\lim }{\left[\frac{f(1+x)}{f\left(1\right)}\right]}^{\frac{1}{x}}$

Considering $y=$${\left[\frac{f(1+x)}{f\left(1\right)}\right]}^{\frac{1}{x}}$

Now taking log both sides

$$\begin{gathered} \Rightarrow \log y=\log {\left[\frac{f(1+x)}{f\left(1\right)}\right]}^{\frac{1}{x}} \\ \Rightarrow \log y=\frac{1}{x}\left[\log f(1+x)-\log f\left(1\right)\right] \end{gathered}$$

Taking limit $x\to 0$, both sides

$$\begin{gathered} \underset{x\to 0}{\lim }\log y=\underset{x\to 0}{\lim }\frac{1}{x}\left[\log f(1+x)-\log f\left(1\right)\right] \\ =\underset{x\to 0}{\lim }\left[\frac{1}{f(1+x)}f\text{'}(1+x)\right][bydefinitionoflimit] \\ =\frac{f\text{'}\left(1\right)}{f\left(1\right)}[ByL\text{'}Hospiyalsrule] \\ =\frac{6}{3}[Accordingtogiveninformation] \\ \log \left[\underset{x\to 0}{\lim }y\right]=2 \\ y=e^2 \end{gathered}$$

Hence, option C is the correct answer.