How to see the order relation for $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$ as $n$ very large? I have two sequences $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$. How do I check whether these two sequence have a fixed order relation when $n$ getting larger? Say, for all $n$ that is large than $N$,  $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}\leq\frac{n^4}{n^7}$.
I know that $\lim_{n\to\infty}\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}=0$, but this doesn't help to see the relation for these two sequence(only helpful to see that the nominator$<$denominator). Also, even I know $$\lim_{n\to\infty}\frac{\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}}{\frac{n^4}{n^7}}=1$$
it appears that we can't deduce which is necessarily bigger than the other for large $n$.
Need help.

Let $n$ be positive and large.  Then
$$
\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2} > \frac{n^4}{n^7}
\\
\Longleftrightarrow
\\
(n^4+99999n^3-99n^2+3)n^7 > n^4(n^7+9999n^5-9999n^2)
\\
\Longleftrightarrow
\\
n^{11}+99999n^{10}-99n^{9}+3n^7 > n^{11}+9999n^{9}-9999n^{6}
\\
\Longleftrightarrow
\\
99999n^{10}-99n^{9}+3n^7 > 9999n^{9}-9999n^{6}
$$
which is true for large $n$.
