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!

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
$$\frac{\sin^2\dfrac x2-\dfrac{x^2}4}{\mathrm e^{x^2}+\mathrm e^{-x^2}-2}\sim_0\frac{-\cfrac{x^4}{48}}{x^4}=-\frac1{48}.$$
