Answer
${e^4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {4x + 1} \right)^{\cot x}} \cr
& {\text{Evaluating the limit when }}x \to {0^ + } \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4x + 1} \right) = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left( {4x + 1} \right)^{\cot x}} = {e^{\cot x\ln \left( {4x + 1} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {4x + 1} \right)^{\cot x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\cot x\ln \left( {4x + 1} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \cot x\ln \left( {4x + 1} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \cot x\ln \left( {4x + 1} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {4x + 1} \right)}}{{\tan x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {4x + 1} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\tan x} \right]}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{4}{{4x + 1}}}}{{{{\sec }^2}x}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{4}{{\left( {4x + 1} \right)\left( {{{\sec }^2}x} \right)}} \cr
& = \frac{4}{{\left( 1 \right)\left( 1 \right)}} = 4 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {e^{\cot x\ln \left( {4x + 1} \right)}} = {e^{\mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 + \frac{1}{x}} \right)}} = {e^4} = {e^L} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {4x + 1} \right)^{\cot x}} = {e^L} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {4x + 1} \right)^{\cot x}} = {e^4} \cr} $$
