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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.

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    $\begingroup$ I believe you only need to apply it $3$ times instead of $10$ times, unless, that is, you're working in ternary, in which you should have written your answer as $10$ $\endgroup$ – Kenny Lau Dec 6 '17 at 20:51
  • $\begingroup$ Maclaurin series? $\endgroup$ – Lord Shark the Unknown Dec 6 '17 at 20:56
  • $\begingroup$ Wonder if there is a way to using that this is $x\frac{f'(x)}{f(x)}$ where $f(x)=\sinh x - \sin x$? $\endgroup$ – Thomas Andrews Dec 6 '17 at 21:09
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$$\begin{array}{cl} & \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\ =& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\ =& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\ =& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\ =& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\ =& 3 \end{array}$$

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Using power series: $$\begin{aligned} \frac{x(\cosh x-\cos x)}{\sinh x-\sin x} &= \frac{x\left((1+\tfrac12 x^2 + O(x^4)) - (1-\tfrac12 x^2 + O(x^4)\right)} {(x+\frac16 x^3 + O(x^5)) - (x - \frac16 x^3 + O(x^5))}\\ &= \frac{x\left(x^2 + O(x^4)\right)} {\frac13 x^3 + O(x^5)}\\ &= \frac{1 + O(x^2)}{\tfrac13 + O(x^2)} = 3 + O(x^2) \end{aligned}$$

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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: $$ \lim_{x\to0} \frac{x(1+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2))} {x+\frac{x^3}{6}-x+\frac{x^3}{6}+o(x^3)} =3 $$

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