Answer
$-\frac{1}{3}$
Work Step by Step
We have
$$
\lim _{x \rightarrow 0}\left(\frac{1}{x^2}- \csc^2 x\right)=\lim _{x \rightarrow 0}\left( \frac{1-x^2\csc}{x^2\sin^2x}\right)=\frac{0}{0}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0}\left( \frac{\sin^2x-x^2}{x^2\sin^2x}\right)=\lim _{x \rightarrow 0}\left( \frac{2\sin x\cos x-2x}{2x\sin^2x+2x^2\sin x\cos x}\right)\\
=\frac{0}{0}.
$$
Which is an intermediate form, so we can simplify the function and then apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow 0}\left( \frac{2\sin x\cos x-2x}{2x\sin^2x+2x^2\sin x\cos x}\right)=\lim _{x \rightarrow 0}\left( \frac{\sin2 x-2x}{2x\sin^2x+x^2\sin2 x}\right)\\
=\lim _{x \rightarrow 0}\left( \frac{2\cos2 x-2}{2\sin^2x+4x\sin x \cos x+2x\sin2 x+2x^2\cos2 x}\right)=\frac{0}{0}.$$
Repeating the above steps as follows, we get:
$$ \lim _{x \rightarrow 0}\left( \frac{2\cos2 x-2}{2\sin^2x+4x\sin x \cos x+2x\sin2 x+2x^2\cos2 x}\right)\\
= \lim _{x \rightarrow 0}\left( \frac{2\cos2 x-2}{2\sin^2x+4x\sin2 x+2x^2\cos2 x}\right)\\
= \lim _{x \rightarrow 0}\left( \frac{-4\sin2x}{4\sin x\cos x+4\sin2 x+8x\cos 2x+4x\cos2 x-4x^2 \sin 2x}\right)=\frac{0}{0}.$$
Repeating the above steps as follows, we get:
$$ \lim _{x \rightarrow 0}\left( \frac{-4\sin2x}{4\sin x\cos x+4\sin2 x+8x\cos 2x+4x\cos2 x-4x^2 \sin 2x}\right)\\
= \lim _{x \rightarrow 0}\left( \frac{-4\sin2x}{6\sin2x+8x\cos 2x+ +4x\cos2 x-4x^2 \sin 2x}\right)\\
= \lim _{x \rightarrow 0}\left( \frac{-8\cos2x}{12\cos2x-16\sin 2x+8\cos2 x-16x\sin 2x+4\cos 2x-8x \sin 2x-8x^2\cos 2x}\right)\\
=-\frac{8}{24}=-\frac{1}{3}$$
