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

\begin{align}
\lim_{x \rightarrow 0} cot(x)-x^{-1} &= \lim_{x \rightarrow 0} \frac{x\cos (x)-\sin(x)}{x\sin(x)} \\
&= \lim_{x \rightarrow 0} \frac{\cos(x)-x\sin(x)-\cos(x)}{\sin(x)+x\cos(x)}, \text{      L'Hôpital's}
\\
&= \lim_{x \rightarrow 0} \frac{-x\sin(x)}{\sin(x)+x\cos(x)}
\\
&= \lim_{x \rightarrow 0} \frac{-\sin(x)-x \cos(x)}{\cos(x)-x\sin(x)+\cos(x)}\text{,      L'Hôpital's}
\\&=0
\end{align}
