I was reading something that mentioned $ \cot {\pi \over 648000} \approx {648000 \over \pi}$ since the argument of $\cot$ is close to zero, and it also said this can be derived by the fact that $\lim_{x \to 0} {\tan x \over x} = 1$.
I simplified the problem as whether
$$\lim_{x \to 0} \left({1 \over \tan x} - {1 \over x}\right) = 0$$
is true, and thought about it, but couldn't figure out the reasoning.
Thank's for a duplicate suggestion, I got to know that this could be solved by applying L'Hospital's Rule twice, but I still wanted to figure out a derivation with $\lim_{x \to 0} {\tan x \over x} = 1$ as was written in the thing I read.
Here's what I tried, getting a hint from this answer.
$$\lim_{x \to 0} \left({1 \over \tan x} - {1 \over x}\right)$$ $$= \lim_{x \to 0} \left({x - \tan x \over x^2} \times {x \over \tan x}\right)$$ $$= \lim_{x \to 0} \left({1 \over x} - {1 \over x} {\tan x \over x}\right)$$
It's really tempting to just replace $\tan x$ with $x$ (which often works to just get an answer) and argue,
$$\lim_{x \to 0} \left({1 \over x} - {1 \over x} {\tan x \over x}\right) = \lim_{x \to 0} \left({1 \over x} - {1 \over x}\right) = 0$$
but that involves distributing $\lim_{x \to 0}$ to $\lim_{x \to 0} {1 \over x}$, which I don't think is good mathematical reasoning.
How can I clean up from the last step?
$$\lim_{x \to 0} \left({1 \over x} - {1 \over x} {\tan x \over x}\right) = 0$$
The Laurent series for the cotangent is $$ \cot x = \frac{1}{x} - \frac{x }{3} - \frac{x^3}{45} - \cdots $$ so for small $x$, $$ \cot x \approx \frac{1}{x}. $$