<s>[INST] Calculate $\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x}$ Beside using l'Hospital 10 times to get 
$$\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} = 3$$ and lots of headaches, what are some elegant ways to calculate the limit?
I've tried to write the functions as powers of $e$ or as power series, but I don't see anything which could lead me to the right result.
 [/INST] $$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\
=& 3
\end{array}$$
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