# 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.

• 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 $10$ – Kenny Lau Dec 6 '17 at 20:51
• Maclaurin series? – Lord Shark the Unknown Dec 6 '17 at 20:56
• Wonder if there is a way to using that this is $x\frac{f'(x)}{f(x)}$ where $f(x)=\sinh x - \sin x$? – Thomas Andrews Dec 6 '17 at 21:09

$$\begin{array}{cl} & \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\ =& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\ =& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\ =& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\ =& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\ =& 3 \end{array}$$
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}
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: $$\lim_{x\to0} \frac{x(1+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2))} {x+\frac{x^3}{6}-x+\frac{x^3}{6}+o(x^3)} =3$$