Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \csc x{\sin ^{ - 1}}x \cr
& {\text{reciprocal identity}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}}} \right){\sin ^{ - 1}}x \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x}}{{\sin x}} \cr
& {\text{evaluating the limit}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x}}{{\sin x}} = \frac{{{{\sin }^{ - 1}}0}}{{\sin 0}} = \frac{0}{0} \cr
& {\text{applying l'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right]}}{{\frac{d}{{dx}}\left[ {\sin x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }}}}{{\cos x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x\sqrt {1 - {x^2}} }} \cr
& {\text{evaluating the limit}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x\sqrt {1 - {x^2}} }} = \frac{1}{{\left( {\cos 0} \right)\left( {\sqrt {1 - {0^2}} } \right)}} \cr
& = 1 \cr} $$
