If $$f\left( x \right) = {\left( {x + 1} \right)^{\cot \,x}}$$    is continuous at $$x = 0,$$  then what is $$f\left( 0 \right)$$  equal to ?

Solution :
For a function to be continuous at a point the limit should be exist and should be equal to the value of the function at the point.
Here point is $$x = 0$$
$$\eqalign{ & {\text{and }}\mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {x + 1} \right)^{\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}.x\,\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {x + 1} \right)^{\frac{1}{x}.\mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan \,x}}}} \cr & = {e^1} \cr & = e \cr} $$
Since, limiting value of $$f\left( x \right) = e,$$   when $$x \to 0,\,f\left( 0 \right)$$   should also be equal to $$e.$$