<s>[INST] 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.
 [/INST] $$\frac{0.5x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\ldots \:}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\ldots \:}}=\frac{0.5+\frac{5}{8}x^2+\frac{23}{144}x^4+\frac{241}{5760}x^6+\frac{3359}{403200}x^{8}+\ldots \:}{{1+\frac{1}{3}x^2+\frac{2}{45}x^4+\frac{1}{315}x^6+\frac{2}{14175}x^{8}+\ldots \:}} \to 1/2$$
as $ x \to 0.$
</s>