(2004) lim_(x to 0)((1)/(sin^2)x-(cos^2x)/(x^2)).

将原式合并为一个分数：
$\lim_{x \to 0}\left(\frac{1}{\sin^2 x} - \frac{\cos^2 x}{x^2}\right) = \lim_{x \to 0}\frac{x^2 - \cos^2 x \sin^2 x}{x^2 \sin^2 x}.$

利用泰勒展开，当 $x \to 0$ 时，$\sin x \approx x - \frac{x^3}{6}$，$\cos x \approx 1 - \frac{x^2}{2}$。代入得：
$\frac{1}{\sin^2 x} \approx \frac{1}{x^2 - \frac{x^4}{3}} \approx \frac{1}{x^2} + \frac{1}{3},$
$\frac{\cos^2 x}{x^2} \approx \frac{1 - x^2}{x^2} = \frac{1}{x^2} - 1.$

相减得：
$\lim_{x \to 0}\left(\frac{1}{x^2} + \frac{1}{3} - \left(\frac{1}{x^2} - 1\right)\right) = \lim_{x \to 0}\left(\frac{1}{3} + 1\right) = \frac{4}{3}.$

答案： $\boxed{\frac{4}{3}}$