Answer
$$5$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}5x}}{x} \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{{{{\sin }^{ - 1}}5\left( 0 \right)}}{0} \cr
& = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so}}{\text{ we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {{{\sin }^{ - 1}}5x} \right)}}{{d/dx\left( x \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{5}{{\sqrt {1 - {{\left( {5x} \right)}^2}} }}}}{1} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{5}{{\sqrt {1 - 25{x^2}} }} \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{5}{{\sqrt {1 - 25{{\left( 0 \right)}^2}} }} \cr
& = 5 \cr} $$
