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 $0.$
You may use $\sin{x} = x+ O(x^3)$ and $\cos{x} = 1+ O(x^2)$:
$$\frac{\sin{x}-x\cos{x}}{x\sin{x}}= \frac{x+ O(x^3) - x(1+O(x^2))}{x(x+ O(x^3))} = \frac{O(x^3)}{x^2(1+ O(x^2))} = \frac{O(x)}{1+ O(x^2)}\stackrel{x\rightarrow 0}{\longrightarrow}0$$
Hint:
$$\lim_{x\to0}\dfrac{\sin x-x\cos x}{x\sin x}=\dfrac1{\lim_{x\to0}\cos x}\cdot\lim_{x\to0}\dfrac{\tan x-x}{x^3}\cdot\dfrac1{\lim_{x\to0}\dfrac{\sin x}x}\cdot\lim_{x\to0}x$$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion
By Taylor’s expansion
$$\frac{1}{x}-\cot x=\frac{1}{x}-\frac1{x+\frac{x^3}3+o(x^3)}=\frac{1+\frac{x^2}3+o(x^2)-1}{x+\frac{x^3}3+o(x^3)}=\frac{\frac{x}3+o(x)}{1+\frac{x^2}3+o(x^2)}\to0$$
Mostly added for your curiosity.
Taylor series make this kind of problem quite simple : they show the limit and how it is approached. Moreover they gave you nice approximations of the function.
Using the classical $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^7\right)$$ $$\frac{\sin{(x)}-x\cos{(x)}}{x\sin{(x)}}=\frac{\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) \right)-x \left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^7\right) \right)}{x\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) \right) }$$ $$\frac{\sin{(x)}-x\cos{(x)}}{x\sin{(x)}}=\frac{\frac{x^3}{3}-\frac{x^5}{30}+O\left(x^7\right) }{x^2-\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^8\right) }=\frac{x}{3}+\frac{x^3}{45}+O\left(x^5\right)$$ Use it for $x=\frac \pi 6$. The exact value is $\frac{6}{\pi }-\sqrt{3}\approx 0.177809$ while the truncated series gives $\frac{\pi \left(540+\pi ^2\right)}{9720}\approx 0.177723$.
