TITLE: Calculate $\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: It's not difficult to show that
$$
\lim_{x\to0}\frac{\cosh x-1}{x^2}=
\lim_{x\to0}\frac{\cosh^2x-1}{x^2(\cosh x+1)}=
\lim_{x\to0}\frac{\sinh^2x}{x^2}\frac{1}{\cosh x+1}=\frac{1}{2}
$$
Similarly,
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}
$$
hence
$$
\lim_{x\to0}\frac{\cosh x-\cos x}{x^2}=1
$$
Therefore your limit is the same as
$$
\lim_{x\to0}\frac{x^3}{\sinh x-\sin x}
$$
If you apply l'Hôpital once, you get
$$
\lim_{x\to0}\frac{3x^2}{\cosh x-\cos x}=3
$$
by the same limit computed before.
With Taylor expansion:
$$
\lim_{x\to0}
\frac{x(1+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2))}
     {x+\frac{x^3}{6}-x+\frac{x^3}{6}+o(x^3)}
=3
$$