$\lim_{x \to 0} \frac {(x^2-\sin x^2) }{ (e^ {x^2}+ e^ {-x^2} -2)}$ solution?

I recently took an math exam where I had this limit to solve

$$\lim_{x \to 0} \frac {(x^2-\sin x^2) }{ (e^ {x^2}+ e^ {-x^2} -2)}$$

and I tought I did it right, since I proceeded like this: 1st I applied Taylor expansion of the terms to the second grade of Taylor, but since I found out the grade in the numerator and in the denominator weren't alike, I chose to try and scale down one grade of Taylor, and I found my self like this:

$$\frac{(x^2-x^2+o(x^2) )}{( (1+x^2)+(1-x^2)-2+o(x^2) )}$$

which should be:

$$\frac{0+o(x^2)}{0+o(x^2)}$$

which should lead to $0$.

Well, my teacher valued this wrong, and I think i'm missing something, I either don't understand how to apply Taylor the right way, or my teacher did a mis-correction (I never was able to see where my teacher said I was wrong, so that's why I'm asking you guys)

Can someone tell me if I really was wrong, and in case I was explain how I should have solved this?

Thanks a lot.

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you weren't allowed to use L'Hopital's Rule? – mathemagician Jan 21 '13 at 17:25
I were allowed to use whatever, but i don't think i grasped it all that well, so that's why i did this wrong... – Yuri Collector's Edition Rossi Jan 21 '13 at 17:30

Your working is correct, except the $o(x^2)$ should be $o(x^6)$ in the numerator and $o(x^4)$ in the denominator. But the main point is that as $x \to 0$, you get $\frac{0}{0}$.

This, however, does not equal to $0$. It's "indeterminate" or not well-defined.

You can consider proceeding by using L'Hopital's Rule, which states that in such cases, the limit does not change if we take derivatives of both numerator and denominator, and still let $x \to 0$.

You may have to use L'Hopital's Rule more than once in some cases, or as already pointed out, factor the expression appropriately before taking derivatives to make the process easier.

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I see, well that's explain why she valued this as wrong... can i ask you to teach me how to apply l'hopital's rule? i have an upcoming make up exam wich should have the same exercise and i yet don't have any knowledge about this rule... or if you can point me some nice source where i could study it and check it in a step-by-step solution, thanks a lot :) – Yuri Collector's Edition Rossi Jan 21 '13 at 17:32
Yuri, yeah sure. The wikipedia article on L'Hopital's Rule has a good first example: en.wikipedia.org/wiki/L'H%C3%B4pital's_rule. Also, if you want something more "blackboard" style and interactive: youtube.com/watch?v=PdSzruR5OeE . All the best. – Conan Wong Jan 21 '13 at 17:34
okay, that video is great (even though i'm not a english native speaker and i thought i would have problem understanding something so hard as math in another language, i did understood it nicely!). so what i had to do was solving everything like i did, and once i had 0/0 i had to go back to the first expression itself and apply l'hopital untill i found a NON indeterminate form? – Yuri Collector's Edition Rossi Jan 21 '13 at 17:47
Yuri, yes, that is correct. Once you have $\frac{0}{0}$ as $x\to 0$, you apply L'Hopital's Rule (sometimes repeatedly, i.e. if the limit is $\frac{0}{0}$ again, take derivatives of top and bottom again) until you find an answer that is not indeterminate. Glad you found the video useful. – Conan Wong Jan 21 '13 at 17:57

$\frac{0}{0}$ is indeterminate-- not $0$. Notice you can factor the bottom (which may make the next step easier):

Apply L'Hospital twice.

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once is enough if you simplify the $2x$ term. – mathemagician Jan 21 '13 at 17:26

$$\lim_{x^2\to0}\frac{x^2-\sin x^2}{e^{x^2}+e^{-x^2}-2}$$ $$=\lim_{y\to0}\frac{y-\sin y}{e^y+e^{-y}-2}$$ $$=\lim_{y\to0}\frac{y-(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots)}{(1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots)+(1-y+\frac{y^2}{2!}-\frac{y^3}{3!}+\cdots)-2}$$

$$=\lim_{y\to0}\frac{\frac{y^3}{3!}-\frac{y^5}{5!}+\cdots}{2(\frac{y^2}{2!}+\frac{y^4}{4!}+\cdots)}$$

$$=\lim_{y\to0}\frac{\frac{y}{3!}-\frac{y^3}{5!}+\cdots}{2(\frac{1}{2!}+\frac{y^2}{4!}+\cdots)}$$ Dividing the numerator & the denominator by $y^2$ as $y\ne 0$ as $y\to0$

So, $$\lim_{y\to0}\frac{y-\sin y}{e^y+e^{-y}-2}=0$$

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isn't that 0/0 too? – Yuri Collector's Edition Rossi Jan 21 '13 at 17:57
@YuriCollector'sEditionRossi, sorry, earlier I excluded the last step – lab bhattacharjee Jan 21 '13 at 18:01
i feel so dumb right now... i should have done this too... thanks btw. – Yuri Collector's Edition Rossi Jan 21 '13 at 18:03

How is $\frac{0+o(x^2)}{0+o(x^2)}$ zero? You need to expand to a degree high enough to keep something nontrivial after cancellation!

Note that $\sin(x^2)=x^2-\frac12 x^4+o(x^6)$ and $e^{\pm x^2}=1+\pm x^2+\frac 12 x^4+o(x^6)$, hence $$f(x)= \frac{\frac12 x^4 + o(x^6)}{x^4+o(x^6)}=\frac{\frac12 + o(x^2)}{1+o(x^2)}\to \frac 12$$

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$\sin y=y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}+\cdots$ right? – lab bhattacharjee Jan 21 '13 at 17:34

It can also be written as ${x^2-\sin x^2}\over{4\sinh^2{x^2/2}}$ $\approx {{x^2-(x^2-x^6/3!)}\over{x^4}}$ $=1/6$

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Re-writing the original expression, the result follows easily
\begin{equation*} \lim_{x\rightarrow 0}\frac{x^{2}-\sin (x^{2})}{e^{x^{2}}+e^{-x^{2}}-2}% =\lim_{t\rightarrow 0}\frac{t-\sin (t)}{e^{t}+e^{-t}-2}=\lim_{t\rightarrow 0}\left( \frac{t-\sin t}{t^{3}}\right) \frac{t}{\left( \frac{e^{t}-1-t}{t^{2}% }\right) +\left( \frac{e^{-t}-1+t}{t^{2}}\right) }=\left( \frac{1}{6}\right) \frac{0}{\left( \frac{1}{2}\right) +\left( \frac{1}{2}\right) }=0. \end{equation*} The key idea in the transformation is to get known expressions. The limits of which can be obtained by L'hospital's rule successively 2 or 3 times.

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