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

Hint:
$$\cot(x)-\frac1x=\frac{\cos(x)}{\sin(x)}-\frac1x=\frac{x\cos(x)-\sin(x)}{x\sin(x)}$$
Now try L'Hospital's rule, as it is now of the form $0/0$.
