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.
 A: $$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\
=& 3
\end{array}$$
A: Using power series:
$$\begin{aligned}
\frac{x(\cosh x-\cos x)}{\sinh x-\sin x}
&= \frac{x\left((1+\tfrac12 x^2 + O(x^4)) - (1-\tfrac12 x^2 + O(x^4)\right)}
{(x+\frac16 x^3 + O(x^5)) - (x - \frac16 x^3 + O(x^5))}\\
&= \frac{x\left(x^2 + O(x^4)\right)}
{\frac13 x^3 + O(x^5)}\\
&= \frac{1 + O(x^2)}{\tfrac13 + O(x^2)} = 3 + O(x^2)
\end{aligned}$$
A: 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:
$$
\lim_{x\to0}
\frac{x(1+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2))}
     {x+\frac{x^3}{6}-x+\frac{x^3}{6}+o(x^3)}
=3
$$
