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$$\begin{align*} \lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\ &= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\ &\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\ &= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\ &\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\ &= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\ \end{align*}$$

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!

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    $\begingroup$ Welcome at MSE. You should use Mathjax to typeset your equations. Mathjax is similar to LaTeX. $\endgroup$ – MrYouMath Sep 24 '17 at 11:25
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    $\begingroup$ L'Hospital's rule is not the alpha and omega of limits computations! $\endgroup$ – Bernard Sep 24 '17 at 11:25
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    $\begingroup$ @MrYouMath Thank you Sir, I am really new to MSE - in fact, I have just created my account to source for help for an assignment. Will take note of your advice for future posts, thank you! $\endgroup$ – Glendon Thaiw Sep 24 '17 at 11:40
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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 24 '17 at 12:05
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    $\begingroup$ If you really want to use LH (have a strong look at Bernard's comment), you need to apply it four times. Just forget it ! It would be a nightmare. $\endgroup$ – Claude Leibovici Sep 24 '17 at 13:45
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Hint: Use the Taylor approximation for sine and the exponential function.

$$\sin(u)=u-u^3/6+O(u^5)$$ $$\exp(u)=1+u+u^2/2 + O(u^3).$$

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$$=\lim_{x\to0}e^{x^2}\cdot\dfrac{\left(\sin\dfrac x2\right)^2-\left(\dfrac x2\right)^2}{(e^{x^2}-1)^2}$$

$$=-\dfrac1{2^4}\lim_{x\to0}e^{x^2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}\cdot\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3}\left(\dfrac1{\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}}\right)^2$$

$\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}=\lim_{x\to0}\left(\dfrac{\sin\dfrac x2}{\dfrac x2}+1\right)=?+1$

$\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}=1$

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 $6I=1$

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  • $\begingroup$ I always love these 'elementary' approaches :-) $\endgroup$ – Simply Beautiful Art Sep 24 '17 at 12:52
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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 $$\frac{\sin^2\dfrac x2-\dfrac{x^2}4}{\mathrm e^{x^2}+\mathrm e^{-x^2}-2}\sim_0\frac{-\cfrac{x^4}{48}}{x^4}=-\frac1{48}.$$

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  • $\begingroup$ I strogly agree with your comment. The result should be $-\frac 1 {48}$. Cheers $\endgroup$ – Claude Leibovici Sep 24 '17 at 13:44
  • $\begingroup$ @ClaudeLeibovici: It took me a hard time to find where the error came from – t shows I'm not very good at typing directly my calculations. Thanks for pointing the error! $\endgroup$ – Bernard Sep 24 '17 at 14:18
  • $\begingroup$ If I was given a dime for any of this kind of mistakes, I should be a millionaire ! $\endgroup$ – Claude Leibovici Sep 24 '17 at 14:22
  • $\begingroup$ Is there a way to be directly in touch ? If you wish, my e-mail address is in my profile. $\endgroup$ – Claude Leibovici Sep 24 '17 at 14:24
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    $\begingroup$ Yes, of course, but I don't want mine to be public, so I'll send you a private e-mail. I finally understood my confusion: rewriting as $\cosh x^2$ and a $-2$ next to it made me think of a cosh squared, not seeing that $\mathrm e^{x^2}\ne\mathrm e^{2x}$. $\endgroup$ – Bernard Sep 24 '17 at 14:30
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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 : $$ \lim_{y\to0} { {\sin^2({\sqrt y \over 2})} - {y \over 4} \over { e^y+ e^{-y}-2} }= \lim_{y\to0} {{{ -{y^2}\over 48}} \over {y^2}} = {-1 \over 48} $$

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  • $\begingroup$ Why do you say clearly denominator is $\infty$? $\endgroup$ – aschepler Sep 24 '17 at 12:46
  • $\begingroup$ Hi Daniel, appreciate the response. Would just like to clarify - by letting y = $x^2$, wouldn't the denominator be 0 as well? Since as y approaches 0, denominator becomes (1+1-2) = 0. $\endgroup$ – Glendon Thaiw Sep 24 '17 at 13:57
  • $\begingroup$ Sorry, i mistake. Both are 0, so need to look in detail using Taylor. $\endgroup$ – Daniel Pol Sep 24 '17 at 13:59
  • $\begingroup$ No problem, I would appreciate if you'd be able to rectify the approach. Thanks! $\endgroup$ – Glendon Thaiw Sep 24 '17 at 14:00
  • $\begingroup$ Thanks for the rectification Daniel, really appreciate it! $\endgroup$ – Glendon Thaiw Sep 24 '17 at 22:00

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