$\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.

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.
