<s>[INST] 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.
 [/INST] If you imagine writing $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}-\frac {n^4}{n^7}$ and putting them over a common denominator, the denominator will be positive.  The first terms in the numerator will be $n^{11}$ and $-n^{11},$ which add to zero.  The next terms will be $99999n^3 \cdot n^7$ and $-9999n^5 \cdot n^4$.  The first is clearly larger, so for large $n$ we have $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}\gt\frac {n^4}{n^7}$  
Added:  another way to look at it, of course with the same result, is to write $$\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}=\frac 1{n^3} \cdot\frac{1+99999n^{-1}-99n^{-2}+3n^{-4}}{1+9999n^{-2}-9999n^{-5}}$$ and use the fact that for $n$ large $\frac 1{1+n^{-1}}\approx 1-n^{-1}$ to make the right side $\frac 1{n^3}(1+99999n^{-1}-9999n^{-2}-99n^{-2}+$ a few other terms)  The leading term is enough to conclude that the fraction goes to zero like $n^{-3}$ but the next term tells you it is greater than $n^{-3}$ for $n$ large enough.  It is just like a Taylor series.  The comparison with $n^{-3}$ subtracts off the leading term, so the next one dominates.
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