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