<s>[INST] 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!
 [/INST] 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
$$\frac{\sin^2\dfrac x2-\dfrac{x^2}4}{\mathrm e^{x^2}+\mathrm e^{-x^2}-2}\sim_0\frac{-\cfrac{x^4}{48}}{x^4}=-\frac1{48}.$$
</s>