Find $\lim\limits_{x \to 0}{\frac{1}{x}(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)})}$. 
$$\lim\limits_{x \to 0}{\frac{1}{x}\left(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)}\right)}$$

I replaced $\sinh(x)$ by $\displaystyle\frac{e^{x}-e^{-x}}{2}$ and used L'Hopital's rule twice but this expression becomes very large. 
Could you please help me

Disclaimer: I am going to write the mother of overkills, just for fun.
From the Weierstrass product for the sine function we have
$$ \frac{x}{\sin(x)} = \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)^{-1} = \prod_{n\geq 1}\left(1+\frac{x^2}{\pi^2 n^2}+\frac{x^4}{\pi^4 n^4}+\ldots\right) \tag{1}$$
and since $\frac{x}{\sin x}$ is an even analytic function in a neighbourhood of the origin we have
$$ \frac{x}{\sin(x)}=1+a_2 x^2+o(x^3),\qquad a_2 = \frac{1}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{\pi^2}\tag{2} $$
and by replacing $x$ with $ix$ we also get
$$ \frac{x}{\sinh(x)}=1-a_2 x^2+o(x^3),\qquad a_2 = \frac{1}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{\pi^2}\tag{3} $$
so:
$$ \lim_{x\to 0}\frac{1}{x}\left(\frac{1}{\sin(x)}-\frac{1}{\sinh x}\right) = \lim_{x\to 0}\frac{1}{x^2}\left(\frac{x}{\sin(x)}-\frac{x}{\sinh x}\right) = 2a_2 = \color{red}{\frac{1}{3}}.\tag{4}$$
