Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \frac{1}{x}} \right) \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \frac{1}{x}} \right) = \frac{1}{{\sin 0}} - \frac{1}{0} = \infty - \infty \cr
& {\text{Where }}\csc x - \frac{1}{x} = \frac{1}{{\sin x}} - \frac{1}{x} = \frac{{x - \sin x}}{{x\sin x}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \sin x}}{{x\sin x}}} \right) \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x\sin x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x\sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {x - \sin x} \right]}}{{\frac{d}{{dx}}\left[ {x\sin x} \right]}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x\cos x + \sin x}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x\cos x + \sin x}} = \frac{0}{0} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {1 - \cos x} \right]}}{{\frac{d}{{dx}}\left[ {x\cos x + \sin x} \right]}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\cos x - x\sin x + \cos x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{2\cos x - x\sin x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{2\cos x - x\sin x}} = \frac{0}{2} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\csc x - \frac{1}{x}} \right) = 0 \cr} $$
