In order that the function f(x) = (x  + 1)^{cot x} is continuous at x = 0, f(0) must be defined as

f(0) = e

For continuity actual value must be equal to limiting value

$A=\lim\limits_{x \rightarrow 0}(x+1)^{\cot x}$

$\log A=\lim\limits_{x \rightarrow 0} \cot x \log (x+1)$

$=\lim\limits_{x \rightarrow 0} \frac{\log (x+1)}{\tan x}$                 $\left[\frac{0}{0} \text { form }\right]$

$=\lim\limits_{x \rightarrow 0} \frac{\frac{1}{x+1}}{\sec ^2 x}=1$             (By L' Hospital Rule)

$\log A=1 \Rightarrow A=e^1=e$

For f(0) must be defined as f(0) = e.

Hence (2) is the correct answer.