Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {x^3}\cot x \cr
& {\text{Rewrite using the reciprocal identity }}\cot x = \frac{1}{{\tan x}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} {x^3}\left( {\frac{1}{{\tan x}}} \right) \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{{x^3}}}{{\tan x}}} \right) \cr
& {\text{Evaluate the limit}} \cr
& = \frac{{{{\left( {{0^ + }} \right)}^3}}}{{\tan \left( {{0^ + }} \right)}} = \frac{0}{0}{\text{ Ind}} \cr
& {\text{Apply L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{{x^3}}}{{\tan x}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {{x^3}} \right]}}{{\frac{d}{{dx}}\left[ {\tan x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3{x^2}}}{{{{\sec }^2}x}} \cr
& {\text{Evaluate the limit}} \cr
& = \frac{{3{{\left( 0 \right)}^2}}}{{{{\sec }^2}\left( {{0^ + }} \right)}} = \frac{0}{1} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^3}\cot x = 0 \cr} $$
