Solve this


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 $x=0]$

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