Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$ 
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$
Here is what I tried:
$$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$
This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here? 
 A: The numerator is 
$$x^2-\frac{x^6}2-x^2\Bigl(1-\frac{x^4}2+\frac{x^8}{24}+o(x^8)\Bigr)=-\frac{x^{10}}{24}+o(x^{10}),$$
hence
$$\frac{x^2-\cfrac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}\sim_0\frac{-\dfrac{x^{10}}{24}}{x^{10}}=-\frac1{24}.$$
*Editors Note: Sorry, can't leave comments yet. When you factor in the $-x^2$ you'll end with $-\frac{1}{24}$.
A: How about factoring by $x^{10}$? 
A: Another way is to note that $\frac{\sin x^{10}}{x^{10}} \to_x 1$, so you get your $x^{10}$ in the denominator. 
A: $$\lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2\cos(x^2)}{\sin(x^{10})} = \lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{24}-\frac{x^{12}}{720}+\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$
$$=\lim_{x\to0}\frac{x^2-\frac{x^6}{2}+(-x^2+\frac{x^6}{2}-\frac{x^{10}}{24}+\frac{x^{12}}{720}-\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$
Canceling terms in the numerator simplifies the equation to:
$$\lim_{x\to0}\frac{-\frac{x^{10}}{24}+\cdots}{x^{10}-\cdots}$$
The other terms in the series are unnecessary, so
$$\lim_{x\to0}(-\frac{x^{10}}{24}*\frac{1}{x^{10}}) = -\frac{1}{24}$$
