$\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}$

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.

Since $\sinh x-\sin x\sim\tfrac13x^3$ while $\cosh x-\cos x\sim x^2$, the limit is $\tfrac13$.

It looks fine, but I would have done it as follows\begin{align}\lim_{x\to0}\frac{\sinh(x)-\sin(x)}{x\bigl(\cosh(x)-\cos(x)\bigr)}&=\lim_{x\to0}\frac{\left(x+\frac{x^3}6+o(x^4)\right)-\left(x-\frac{x^3}6+o(x^4)\right)}{x\left(\left(1+\frac{x^2}2+o(x^3)\right)-\left(1-\frac{x^2}2+o(x^3)\right)\right)}\\&=\lim_{x\to0}\frac{\frac{x^3}3+o(x^4)}{x+o(x^4)}\\&=\lim_{x\to0}\frac{\frac13+o(x^3)}{1+o(x^3)}\\&=\frac13.\end{align}