Is it possible to evaluate $$\lim_{x\to 0} \left(\frac{1}{x}-\cot x\right) $$ without L'Hopital's rule? The most straightforward way is to use $\sin{x}$ and $\cos{x}$ and apply the rule, but I stuck when I arrived to this part (since I don't want to use the rule as it is pretty much cheating):
$$\lim_{x\to0} \left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right) $$
It seems like there's something to do with the identity
$$\lim_{x\to0} \frac{\sin{x}}{x}=1$$
but I can't seem to get the result of $0.$
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