Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \cot x} \right) \cr
& {\text{Evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \cot x} \right) = \csc 0 - \cot 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \cot x} \right) = \infty - \infty \cr
& {\text{Use the basic trigonometric identities}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right) \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{{\sin x}}} \right) \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{1 - \cos 0}}{{\sin 0}} = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so }}{\text{we apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {1 - \cos x} \right)}}{{d/dx\left( {\sin x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\cos x}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{\sin 0}}{{\cos 0}} \cr
& = 0 \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \cot x} \right) = 0 \cr} $$
