Answer
$$e$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + x} \right)^{\cot x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + x} \right)^{\cot x}} = {\left( {1 + 0} \right)^{\cot 0}} = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {\text{Nothing that }}{\left( {1 + x} \right)^{\cot x}} = {e^{\cot x\ln \left( {1 + x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + x} \right)^{\cot x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\cot x\ln \left( {1 + x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \cot x\ln \left( {1 + x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \cot x\ln \left( {1 + x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {1 + x} \right)}}{{\frac{1}{{\cot x}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {1 + x} \right)}}{{\tan x}} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {1 + x} \right)}}{{\tan x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{1}{{1 + x}}}}{{{{\sec }^2}x}} = \frac{{\frac{1}{{1 + 0}}}}{{{{\sec }^2}0}} = 1 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + x} \right)^{\cot x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\cot x\ln \left( {1 + x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \cot x\ln \left( {1 + x} \right)}} = {e^1} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + x} \right)^{\cot x}} = e \cr} $$
