[Math] limit of $x \cot x$ as $x\to 0$.

Question

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 1

HINT: rewrite your term in the form $$\frac{\cos(x)}{\frac{\sin(x)}{x}}$$