TITLE: How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$?
QUESTION [5 upvotes]: $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
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!

REPLY [2 votes]: $$=\lim_{x\to0}e^{x^2}\cdot\dfrac{\left(\sin\dfrac x2\right)^2-\left(\dfrac x2\right)^2}{(e^{x^2}-1)^2}$$
$$=-\dfrac1{2^4}\lim_{x\to0}e^{x^2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}\cdot\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3}\left(\dfrac1{\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}}\right)^2$$
$\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}=\lim_{x\to0}\left(\dfrac{\sin\dfrac x2}{\dfrac x2}+1\right)=?+1$
$\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}=1$
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 $6I=1$