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?
 A: As $\;x\to\infty\;$ :
$$\frac{x^2e^{x^4}-\sin x^2}{1-\cos x^3}\ge\frac{x^2e^{x^4}-1}2\xrightarrow[x\to\infty]{}\infty$$
A: From here you with some adjustment conclude:
$$...=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6++\mathcal{O}(x^{10})}{\frac{1}{2}x^6+\mathcal{O}(x^{10})}=\frac73+\mathcal{O}(x^{4})\to\frac73$$
A: 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
A: $\mathcal{O}(x^n)$ stands for an “unnamed” function $f(x)$ such that there exists a positive constant $M$ so that $|f(x)|\le M|x^n|$ for $x$ sufficiently close to $0$. In particular, $f(x)$ can be a function such that $\lim_{x\to0}f(x)/x^n$ is finite.
As a consequence, $\mathcal{O}(x^n)$ is also $\mathcal{O}(x^m)$ if $m\le n$: if you can find $M$ as before, then
$$
|f(x)|\le M|x^n|=M|x^m|\,|x^{n-m}|\le M|x^m|
$$
as soon as $|x|<1$.
Also needed here is that, if $f(x)$ is $\mathcal{O}(x^n)$, then $x^kf(x)$ is $\mathcal{O}(x^{n+k})$.
By the triangle inequality, the sum of two functions that are $\mathcal{O}(x^n)$ is also a function that is $\mathcal{O}(x^n)$.
For your numerator you can write
$$
e^{x^4}=1+x^4+\mathcal{O}(x^8)
$$
and so
$$
x^2e^{x^4}=x^2+x^6+\mathcal{O}(x^{10})
$$
Therefore
$$
x^2e^{x^4}=x^2+x^6-x^2+\frac{x^6}{6}+\mathcal{O}(x^{10})=\frac{7}{6}x^6+\mathcal{O}(x^{10})
$$
The denominator is
$$
1-1+\frac{1}{2}x^6+\mathcal{O}(x^{12})=\frac{1}{2}x^6+\mathcal{O}(x^{10})
$$
Finally, you can prove that if $n>m$ and $f(x)$ is $\mathcal{O}(x^n)$, then $f(x)/x^m$ is $\mathcal{O}(x^{n-m})$.

You may want to use the “small o” notation; roughly speaking, $o(x^n)$ means you disregard terms with order larger than $x^n$. In this case
$$
x^2e^{x^4}-\sin(x^2)=
x^2(1+x^4+o(x^4))-\left(x^2-\frac{1}{6}x^6+o(x^6)\right)=\frac{7}{6}x^6+o(x^6)
$$
and
$$
1-\cos(x^3)=1-1+\frac{1}{2}x^6+o(x^6)
$$
so your limit becomes
$$
\lim_{x\to0}\frac{\frac{7}{6}x^6+o(x^6)}{\frac{1}{2}x^6+o(x^6)}
=
\lim_{x\to0}\frac{\frac{7}{6}+o(x^0)}{\frac{1}{2}+o(x^0)}
$$
and you're done because $\lim_{x\to0}o(x^0)=0$.
