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
 A: HINT:
Instead of using L'Hospital's Rule, we can apply Taylor's Theorem and obtain
$$\begin{align}
\frac1x\left(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)}\right)&=\frac1x\left(\frac{\sinh(x)-\sin(x)}{\sin(x)\sinh(x)}\right)\\\\
&=\frac{\left(x+\frac16x^3+O(x^5)\right)-\left(x-\frac16x^3+O(x^5)\right)}{x^3+O(x^5)}
\end{align}$$
Can you finish now?
A: write your term in the form $$\lim_{x\to 0}\frac{\sinh(x)-\sin(x)}{x\sin(x)\sinh(x)}$$ and use L'Hospital
A: A little bit of algebra:
$$\begin{align}
\frac1x\left(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)}\right)&=\frac1x\left(\frac{\sinh(x)-\sin(x)}{\sin(x)\sinh(x)}\right)\\\\
&={x \over \sin x}{x \over \sinh x}{\sinh x -x+x-\sin x \over x^3}\\\\
&={x \over \sin x}{x \over \sinh x}\left ({\sinh x -x \over x^3}+{x-\sin x \over x^3}\right )
\end{align}$$
All the factors have a finite limit, so you can substitute each with its limit and find the answer you need.
A: 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:
$$ \lim_{x\to 0}\frac{1}{x}\left(\frac{1}{\sin(x)}-\frac{1}{\sinh x}\right) = \lim_{x\to 0}\frac{1}{x^2}\left(\frac{x}{\sin(x)}-\frac{x}{\sinh x}\right) = 2a_2 = \color{red}{\frac{1}{3}}.\tag{4}$$
A: hint
for numerator,
$$\sinh (x )=x+\frac {x^3}{6}(1+\epsilon_1 (x)) $$
$$\sin (x)=x-\frac {x^3}{6}(1+\epsilon_2 (x)) $$
$$\sinh (x)-\sin (x)\sim \frac {2x^3}{6} (x\to 0) $$
your limit is $$\frac {1}{3} .$$
for denominator, use
$$\sin (x)\sim x (x\to 0) $$ and
$$\sinh (x)\sim x (x\to 0) $$
A: Note that 
$$\frac{1}{x}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\frac{x}{\sin x}\frac{x-\sin x}{x^3}$$ is easy to evaluate. 
Now you try 
$$\frac{1}{x}\left(\frac{1}{x}-\frac{1}{\sinh x}\right).$$
