If $f(x)=(x+1)^{\cot x}$ be continuous at $x=0$, then $f(0)$ is equal to
(a) 0
(b) $1 / e$
(c) $\mathrm{e}$
(d) none of these
(c) $e$
Suppose $f(x)$ is continuous at $x=0$.
Given: $f(x)=(x+1)^{\cot x}$
$\log f(x)=(\cot x)(\log (x+1)) \quad[$ Taking $\log$ on both sides $]$
$\Rightarrow \lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0}(\cot x)(\log (x+1))$
$\Rightarrow \lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0}\left(\frac{\log (x+1)}{\tan x}\right)$
$\Rightarrow \lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0} \frac{\left(\frac{\log (x+1)}{x}\right)}{\left(\frac{\tan x}{x}\right)}$
$\Rightarrow \lim _{x \rightarrow 0} \log f(x)=\frac{\lim _{x \rightarrow 0}\left(\frac{\log (x+1)}{x}\right)}{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)}$
$\Rightarrow \log \left(\lim _{x \rightarrow 0} f(x)\right)=\frac{\lim _{z \rightarrow 0}\left(\frac{\log (x+1)}{z}\right)}{\lim _{z \rightarrow 0}\left(\frac{\tan x}{x}\right)}$ $[\because f(x)$ is continuous at $x=0]$
$\Rightarrow \log \left(\lim _{x \rightarrow 0} f(x)\right)=1$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=e$
$\Rightarrow f(0)=e \quad[\because f(x)$ is continuous at $x=0]$