How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.
However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants. 
In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.
Any help would be appreciated, thank you all!
 A: Hint: Use the Taylor approximation for sine and the exponential function.
$$\sin(u)=u-u^3/6+O(u^5)$$
$$\exp(u)=1+u+u^2/2 + O(u^3).$$
A: $$=\lim_{x\to0}e^{x^2}\cdot\dfrac{\left(\sin\dfrac x2\right)^2-\left(\dfrac x2\right)^2}{(e^{x^2}-1)^2}$$
$$=-\dfrac1{2^4}\lim_{x\to0}e^{x^2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}\cdot\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3}\left(\dfrac1{\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}}\right)^2$$
$\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}=\lim_{x\to0}\left(\dfrac{\sin\dfrac x2}{\dfrac x2}+1\right)=?+1$
$\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}=1$
For $I=\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3},$ set $x=2y$
and use Are all limits solvable without L'Hôpital Rule or Series Expansion, to find $6I=1$
A: You only have to find equivalents for the numerator and the denominator. We'll begin with rewriting them, and use classical Taylor's expansions:
Numerator:
 $$\sin^2\dfrac x2-\dfrac{x^2}4=\dfrac{1-\cos x}{2}-\dfrac{x^2}4=\biggl[\frac12-\Bigl(\frac12-\frac{x^2}{4}+\frac{x^4}{48}+o\bigl(x^4\bigr)\Bigr)\biggr]-\frac{x^2}4=-\frac{x^4}{48}+o\bigl(x^4\bigr)$$
so the numerator is equivalent near $\,0\;$ to $-\dfrac{x^4}{48}$.
Denominator:
We know $\sinh u\sim_0 u$, so 
$$\mathrm e^{x^2}+\mathrm e^{-x^2}-2=1+x^2+\frac{x^4}2+o(x^4)+1-x^2+\frac{x^4}2+o(x^4)-2=x^4+o(x^4)$$
so the denominator is equivalent to $x^4$. There results that
$$\frac{\sin^2\dfrac x2-\dfrac{x^2}4}{\mathrm e^{x^2}+\mathrm e^{-x^2}-2}\sim_0\frac{-\cfrac{x^4}{48}}{x^4}=-\frac1{48}.$$
A: 
I think, from the second expression you will get easy a result : 
$$
\lim_{x\to0} { {\sin^2({x \over 2})} - {{x^2} \over 4} \over 
{ e^{x^2}+ e^{-{x^2}}-2} } = \lim_{y\to0} { {\sin^2({\sqrt y \over 2})} - {y \over 4} \over 
{ e^y+ e^{-y}-2} }=?
$$
I do $y=x^2$. Clearly numerator is 0 and denominator is $0$. 
I will correct here. I expand function $\sin^2{y \over 2}={y \over 4} - {{y^2} \over 48}+ ...$. So, at numerator we get ${{-{y^2}}\over 48}+ ...$. At denominator a Taylor expansion get after resting 2 : $y^2 + ...$
We do a division :
$$
\lim_{y\to0} { {\sin^2({\sqrt y \over 2})} - {y \over 4} \over 
{ e^y+ e^{-y}-2} }= \lim_{y\to0} {{{ -{y^2}\over 48}} \over {y^2}} = {-1 \over 48}
$$
