38.3. \( \lim _{x \rightarrow 0}\left(\frac{\cos ^{2} x-1}{x \sin x}\right) \)

To evaluate the limit \( \lim _{x \rightarrow 0}\left(\frac{\cos ^{2} x-1}{x \sin x}\right) \), we first simplify the expression in the limit. Recall the identity \( \cos^2 x - 1 = -\sin^2 x \). Thus, we can rewrite the limit as follows: \[ \lim _{x \rightarrow 0}\left(\frac{-\sin^2 x}{x \sin x}\right) = \lim _{x \rightarrow 0}\left(-\frac{\sin^2 x}{x \sin x}\right) \] This can be further simplified to: \[ = \lim _{x \rightarrow 0}\left(-\frac{\sin x}{x}\right) \cdot \sin x \] As \( x \) approaches 0, we know that \( \frac{\sin x}{x} \) approaches 1, and \( \sin x \) approaches 0. Therefore, we can substitute these limits into our expression: \[ = -1 \cdot 0 = 0 \] Thus, the limit is \[ \boxed{0} \]