Let us consider the following limit:
$$ \lim\limits_{x \to 0} \frac{\sin(x^2)-x+\frac{x^6}{6}-x^{10}}{x^{10}} $$
I use the Taylor's series for $\sin(x^2)$ and, by doing computations I have $$ \lim\limits_{x \to 0} \frac{x^2-x-\frac{119}{120}x^{10}+o(x^{10})}{x^{10}} $$
The solution I have doesn't explain it, but asserts that the limit is exactly $-\frac{119}{120}$. May I use the fact that the numerator is asymptotically equivalent to $-\frac{119}{120}x^{10}$ in order to reduce it? It seems really strange. Or this limit goes to $\infty$, as I expect?
We have that $$\sin(x^2)-x+\frac{x^6}{6}-x^{10}=-x+o(x).$$ and the given limit is an undetermined infinity $$\lim\limits_{x \to 0^{\pm}} \frac{\sin(x^2)-x+\frac{x^6}{6}-x^{10}}{x^{10}}=\mp\infty.$$ I guess that your limit is $$\lim\limits_{x \to 0} \frac{\sin(x^2)-x^2+\frac{x^6}{6}-x^{10}}{x^{10}}= \lim\limits_{x \to 0} \frac{\frac{x^{10}}{5!}-x^{10}+o(x^{10})}{x^{10}}=\frac{1}{5!}-1=\frac{119}{120}.$$