Trouble calcualting simple limit I'm not sure if my solution is correct. The limit is:
$\lim_{x\to0}\cot(x)-\frac{1}{x}$. Here is how I tried to solve it:
 1. $\lim_{x\to0}\cot(x)-\frac{1}{x}$ = $\lim_{x\to0}\cot(x) - x^{-1}$
 2. Since $\cot(0)$ is not valid, apply the de l'Hôpital rule  $(\cot (x) )' = -\frac{1}{\sin^2 x}=\sin^{-2} x $ and $(x^{-1})'$ = $x^{-2}$ 
 3. $\lim_{x\to0}\sin^{-2} (x)-x^{-2} = 0 - 0 = 0$ 
However I'm not sure that my logic is correct

For the first one, $\tan x \sim x$ for small $x$ by Taylor's yields $\cot x\sim 1/x$ which should make the limit pretty easy.
To apply L'Hôpital's I always convert it to the indeterminate form $\frac{\infty}{\infty}$ to avoid mistakes. Although you are justified in applying L'Hôpital's in this case as $\infty-\infty$ is an indeterminate form.
