As stated in the title.
My attempt, begin with L'Hopital:
$$L=\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}=\lim_{x \to 0}{\frac{\cosh(x)-\cos(x)}{(\cosh(x)-\cos(x))+x(\sinh(x)+\sin(x))}}$$
Dividing by the numerator
$$\lim_{x \to 0}\frac{1}{1+\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}=\frac{1}{1+\lim_{x \to 0}{\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}}$$
L'Hopital again
$$\frac{1}{1+\lim_{x \to 0}{\left(\frac{x(\cosh(x)+\cos(x))+(\sinh(x)+\sin(x))}{\sinh(x)+\sin(x)}\right)}}$$
Diving through
$$\frac{1}{1+\lim_{x \to 0}{\left(1+\frac{\cosh(x)+\cos(x)}{\left(\frac{\sinh(x)}{x}+\frac{\sin(x)}{x} \right)}\right)}}=\frac{1}{1+(1+\frac{1+1}{1+1})}=\frac{1}{3}$$
Is this correct, and is there a more elegant way of doing it?
Best Answer
Hint: In fact, \begin{eqnarray} L&=&\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}\\ &=&\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x^3}}\cdot\lim_{x \to 0}{\frac{x^2}{\cosh(x)-\cos(x)}}\\ &=&\lim_{x \to 0}{\frac{\cosh(x)-\cos(x)}{3x^2}}\cdot\lim_{x \to 0}{\frac{2x}{\sinh(x)+\sin(x)}} \end{eqnarray} and you can continue.