limit of $x \cot x$ as $x\to 0$.

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?

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.
HINT: rewrite your term in the form $$\frac{\cos(x)}{\frac{\sin(x)}{x}}$$
You could have stopped at $$\lim_{x\to0}\frac{\cos(x)-x\sin(x)}{\cos(x)}=\frac11$$ but otherwise, your answer is fine.
If you can use $$\lim_{x\to0}\frac{\sin(x)}x=1$$ it would simplify the computation as well.
Notice,$$\lim_{x\to 0}xcotx$$ $$=\lim_{x\to 0}\left(\frac{x}{\tan x}\right)=1$$ Or using L-Hospital's rule for $\frac{0}{0}$ form as follows $$\lim_{x\to 0}\left(\frac{x}{\tan x}\right)=\lim_{x\to 0}\frac{\frac{d(x)}{dx}}{\frac{d}{dx}(\tan x)}$$ $$=\lim_{x\to 0}\frac{1}{\sec^2x}=\frac{1}{\sec^2(0)}=\frac{1}{1}=1$$