calculus - How find this limit $lim_{xto 0}frac{1}{x^4}left(frac{1}{x}left(frac{1}{tanh{x}}-frac{1}{tan{x}}right)-frac{2}{3}right)=？$

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

Answer

$$\coth x=\frac{1}{x}+\frac{x}{3}-\frac{x^3}{45}+\frac{2x^5}{945}-\frac{x^7}{4725}+... \\ \cot x =\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\frac{x^7}{4725}-... \\[0.9cm] \text{Given limit is}\\ \lim_{x \to 0} \frac{1}{x^4}\left(\frac{1}{x}\left(\frac{2x}{3}+\frac{4x^5}{945} + ...\right)-\frac{2}{3}\right) \\ = \lim_{x \to 0} \frac{1}{x^4}\left(\frac{4x^4}{945} + ...\right) \\ =\frac{4}{945}$$

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.