Limit question involving L'Hospitals rule I need help in solving the below question using L'Hospitals rule:
$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) $$
I'm getting infinity as the final answer. Thanks in advance.

$$
\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) \\
= \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\frac{\cos^2 x}{\sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x}{x^2 \sin^2 x}-\frac{x^2 \cos^2 x}{x^2 \sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x - x^2 \cos x}{x^2 \sin^2 x} \right) \\
$$
Now you have a fraction where both numerator and denominator approach $0$ as $x \rightarrow 0$, so you can invoke L'Hospital's rule by taking the derivative of both top and bottom.
