Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$ 
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$

By taylor polynomials we get: $e^{x^4}=1+x^4+\frac{x^8}{2}+\mathcal{O}(x^{12})$
$\sin(x^2)=x^2-\frac{x^6}{6}+\mathcal{O}(x^{10})$
$\cos(x^3)=1-\frac{x^6}{2}+\mathcal{O}(x^{12})$
so putting these together:
$$  \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)} = \frac{x^2+x^6+\frac{x^{10}}{2}-x^2+\frac{x^6}{6}-\mathcal{O}(x^{10})}{\frac{x^6}{2}-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}$$
Now I am not too familiar with the Big-Oh notation for limits so I am stuck here.
How does arithmetic work with them, can I simplify the oh's in the numerator and can I take $x$'s out?

Assuming the limr $x\to0$
$$\dfrac{\lim_{x\to0}\dfrac{e^{x^4}-1}{x^4}+\lim_{x\to0}\dfrac{x^2-\sin(x^2)}{x^6}}{\left(\lim_{x\to0}\dfrac{\sin x^3}{x^3}\right)^2}\cdot\lim_{x\to0}(1+\cos x^3)$$
$$=\dfrac{\left(1+\dfrac16\right)(1+\cos0)}{1^2}$$
using Are all limits solvable without L'Hôpital Rule or Series Expansion
