\( \begin{array}{l} \text { Use L'hopital's to evaluate }\\ \lim _{x \rightarrow 0} \sin ^{-1} x \csc x, \sin ^{-1} 0=0 \end{array} \)

To evaluate the limit \( \lim _{x \rightarrow 0} \sin^{-1} x \csc x \), we can rewrite \(\csc x\) as \(\frac{1}{\sin x}\), so the limit becomes \( \lim_{x \to 0} \frac{\sin^{-1} x}{\sin x} \). As \(x\) approaches 0, both the numerator and the denominator approach 0, creating a \( \frac{0}{0} \) indeterminate form. Now we can apply L'Hôpital's Rule. Taking the derivatives, we find \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \) and \( \frac{d}{dx}(\sin x) = \cos x \). Thus, the limit becomes: \[ \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 - x^2}}}{\cos x} \] Substituting \(x = 0\) gives: \[ \frac{\frac{1}{\sqrt{1 - 0^2}}}{\cos(0)} = \frac{1}{1} = 1. \] Therefore, the limit evaluates to \(1\).