calculus - limit of $x cot x$ as $xto 0$.

I was asked to calculate

$$\lim_{x \to 0}x\cot x$$
I did it as following (using L'Hôpital's rule):
$$\lim_{x\to 0} x\cot x = \lim_{x\to 0} \frac{x \cos x}{\sin x}$$ We can now use L'Hospital's rule since the limit has indeterminate form $\frac{0}{0}$. Hence $$\begin{align}\lim_{x\to 0}\frac{(x \cos x)'}{(\sin x)'} &= \lim_{x\to 0}\frac{-x\sin x + \cos x}{\cos x} \\ &= \lim_{x\to 0}\frac{-x\sin x}{\cos x} + 1 \\[4pt ]&= \lim_{x\to 0} - x \tan x + 1 \\[4pt] &= 1 \end{align}$$
I think that the result is correct but are the arguments correct?

Answer

Welcome to MSE! As best as I can tell your reasoning is sound, although as other users pointed out you could have done this in a few less steps. Regardless, your work is clear and your answer is correct. On an aside I am happy to see that you formatted your question with $\LaTeX$. I added some additional formatting to clean things up a bit, but overall nicely done.