A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$ I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.
This is what I got
$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.

You may use some standard limits as follows:
\begin{eqnarray*}\frac{e^{x^2}-\cos x - x\sin x}{\sinh^2 x}
& = & \left(\frac{e^{x^2}-1}{x^2}+ \frac{1-\cos x}{x^2}-\frac{\sin x}{x}\right)\cdot \left(\frac{x}{\sinh x}\right)^2\\
& \stackrel{x\to 0}{\longrightarrow} & \left(1+\frac 12 - 1\right)\cdot 1^2 = \frac 12
\end{eqnarray*}
