How find this limit $\lim_{x\to 0}\frac{1}{x^4}\left(\frac{1}{x}\left(\frac{1}{\tanh{x}}-\frac{1}{\tan{x}}\right)-\frac{2}{3}\right)=？$ Find this following limit
$$\displaystyle \lim_{x\to 0}\dfrac{1}{x^4}\left(\dfrac{1}{x}\left(\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}\right)-\dfrac{2}{3}\right)=？$$

My try: since $$\tanh{x}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}$$
then
$$\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}-\dfrac{1}{\tan{x}}$$

then I can't,Thank you
$$
\coth x=\frac{1}{x}+\frac{x}{3}-\frac{x^3}{45}+\frac{2x^5}{945}-\frac{x^7}{4725}+... \\
\cot x =\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\frac{x^7}{4725}-... \\[0.9cm]
\text{Given limit is}\\ \lim_{x \to 0} \frac{1}{x^4}\left(\frac{1}{x}\left(\frac{2x}{3}+\frac{4x^5}{945} + ...\right)-\frac{2}{3}\right) \\
= \lim_{x \to 0} \frac{1}{x^4}\left(\frac{4x^4}{945} + ...\right) \\
=\frac{4}{945}
$$

Don't worry if you can't remember these power series. You can derive them in a jiffy using Bernoulli numbers. But that's for another day.
Take the power series expansion of tanh to fifth order, invert it.  Then take power series expanion of tan to fifth order, invert it.  Take the difference.  The rest is clear sailing. Note that the expansions will only have 3 nonzero terms since the functions are odd.