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.
 A: Hint
$$e^{x^2}-\cos(x)-x\sin(x)=\frac{x^2}{2}+o(x^2)$$
and
$$\sinh(x)^2=x^2+o(x^2).$$
A: $$\frac{0.5x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\ldots \:}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\ldots \:}}=\frac{0.5+\frac{5}{8}x^2+\frac{23}{144}x^4+\frac{241}{5760}x^6+\frac{3359}{403200}x^{8}+\ldots \:}{{1+\frac{1}{3}x^2+\frac{2}{45}x^4+\frac{1}{315}x^6+\frac{2}{14175}x^{8}+\ldots \:}} \to 1/2$$
as $ x \to 0.$
A: 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*}
A: Here's an elementary solution,(I mean elementary as I have expressed the limit in question, to various standard limits that a beginner in calculus faces.)
The limit in question is taken to be $L$
$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)+(1-\cos x)-x\sin x}{\sinh^2x}$
$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)}{\sinh^2x}+\frac{(1-\cos x)}{\sinh^2x}-\frac{x\sin x}{\sinh^2x}$
$$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)}{x^2}\frac{x^2}{\sinh^2x}+\frac{2\sin^2\left(\frac{x^2}{2}\right)}{\sinh^2x}-\frac{x\sin x}{\sinh^2x}=1+\frac{1}{2}-1=\frac{1}{2}$$
A: It is possible to use the Hopital rule even though, in order to obtain the result, it has to be used twice. For sake of personal preference I will expand the function $\sinh x$ as a sum of exponential functions.
$$\sinh^{2}x=\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}=\frac{e^{2x}+e^{-2x}-2}{4}$$
By substitution, the following limit is obtained where, in the second step, the 4 has been factored out from the limit.
$$\lim_{x \to 0} \frac{e^{x^{2}}-\cos x-x\sin x}{\sinh^{2}x}=4\lim_{x \to 0} \frac{e^{x^{2}}-\cos x-x\sin x}{e^{2x}+e^{-2x}-2}=\left[\frac{0}{0}\right]$$
By obtaining an indeterminate form such as $\left[\frac{0}{0}\right]$ it is possible to use the Hopital rule, so by differentiating the numerator and the denominator we have:
$$4\lim_{x \to 0}\frac {2xe^{x^{2}}+\sin x-\sin x-x\cos x}{2e^{2x}-2e^{-2x}}=\left[\frac{0}{0}\right]$$
Reiterating the application of the same principle stated above we the following result is obtained.
$$4\lim_{x \to 0}\frac{2e^{x^{2}}+4x^{2}e^{x^{2}}+x\sin x-\cos x}{4e^{2x}-4e^{-2x}}=4\frac{2-1}{4+4}=\frac{1}{2}$$
