Is there a faster way to solve this limit, $\displaystyle \displaystyle\lim_{x\to{0}}\frac{1}{x^2}-cot^2\;x$ other than using L'Hopital's Rule twice?
You can always fall back on the Taylor series
The first few terms is about all you need.
$\displaystyle cot^{2}(x)=\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)$
Now, you have:
$\displaystyle \frac{1}{x^{2}}-\left(\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)\right)$
$\displaystyle \lim_{x\rightarrow{0}}\frac{2}{2x^{2}+3}=\frac{2}{ 3}$
I agree with Jhevon, and it is also more interesting if you try to avoid using L' Hôpital.
A bit about Taylor polynomials: the nth degree polynomial of $\displaystyle f(x)$, $\displaystyle T_n$, centered in , let's say "a", has the following property: $\displaystyle T_n(a)=f(a)$ , $\displaystyle T_n'(a)=f'(a)$ ... $\displaystyle T_n^{(n)}(a)=f^{(n)}(a)$.
It turns out to be a good approximation of the function (in a certain interval)
The remainder is given by $\displaystyle f(x)-T_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}\cdot{(x-a)^{n+1}}$ for some $\displaystyle c\in(a, x)$
You can check that: $\displaystyle T_n=f(a)+\sum_{k=1}^n{\left(\frac{f^{(k)}(a)}{k!}\ cdot{(x-a)^k}\right)}$
Look it up in your calculus book, it should have something about this