$\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.

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
$$f(x)= \frac{\frac12 x^4 + o(x^6)}{x^4+o(x^6)}=\frac{\frac12  + o(x^2)}{1+o(x^2)}\to \frac 12$$
