Question

If f (x) = (x + 1)^{cot x} be continuous at x = 0, then f (0) is equal to
(a) 0
(b) 1/e
(c) e
(d) none of these

Answer 1

(c) e

Suppose $f\left(x\right)$ is continuous at $x=0.$

Given: $f\left(x\right)={\left(x+1\right)}^{\cot x}$

$$\begin{gathered} \log f\left(x\right)=\left(\cot x\right)\left(\log \left(x+1\right)\right)\left[\mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow \underset{x\to 0}{\lim }\log f\left(x\right)=\underset{x\to 0}{\lim }\left(\cot x\right)\left(\log \left(x+1\right)\right) \\ \Rightarrow \underset{x\to 0}{\lim }\log f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{\log \left(x+1\right)}{\tan x}\right) \\ \Rightarrow \underset{x\to 0}{\lim }\log f\left(x\right)=\underset{x\to 0}{\lim }\frac{\left({\displaystyle \frac{\log \left(x+1\right)}{x}}\right)}{\left({\displaystyle \frac{\tan x}{x}}\right)} \\ \Rightarrow \underset{x\to 0}{\lim }\log f\left(x\right)=\frac{\underset{x\to 0}{\lim }{\displaystyle \left(\frac{\log \left(x+1\right)}{x}\right)}}{\underset{x\to 0}{\lim }{\displaystyle \left(\frac{\tan x}{x}\right)}} \\ \Rightarrow \log \left(\underset{x\to 0}{\lim }f\left(x\right)\right)=\frac{\underset{x\to 0}{\lim }{\displaystyle \left(\frac{\log \left(x+1\right)}{x}\right)}}{\underset{x\to 0}{\lim }\left({\displaystyle \frac{\tan x}{x}}\right)}\left[\because f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0\right] \\ \Rightarrow \log \left(\underset{x\to 0}{\lim }f\left(x\right)\right)=1 \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=e \\ \Rightarrow f\left(0\right)=e\left[\because f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0\right] \end{gathered}$$