Proving a Limit I need to prove that
$$\displaystyle\lim_{n→∞} \frac{4n^5 − 1000}{2n^5 − n^3 + 2000} = 2$$
$n$ is a natural number.
I know that I need to use 
$$∀ε > 0 : ∃N : n ≥ N ⇒ |(4n^5-1000/2n^5-n^3+200) − 2| <ε$$
I'm just not sure where exactly to go from here. 
 A: You already know what to do, so I'll give you a hint that will make it easier. Try this:
$$\dfrac{4n^5 − 1000}{2n^5 − n^3 + 2000}=\dfrac{(4n^5-2n^3 +4000)+(2n^3 -4000 -1000)}{2n^5-n^3+2000}= 2 +\dfrac{2n^3 -5000}{2n^5-n^3+2000} $$
A: For sufficiently large $n$
$\frac{4n^5 − 1000}{2n^5 − n^3 + 2000}= \frac {4n^5 -2n^3 +4000}{2n^5 − n^3 + 2000}+ \frac {2n^3 - 5000}{2n^5 − n^3 + 2000}=$
$2 + \frac{2}{2n^2 - 1 + 2000/n^3} - \frac {5000}{2n^5 - n+ 2000}$
You can use whatever method you want to show the limits of $ \frac{2}{2n^2 - 1 + 2000/n^3}$ and $\frac {5000}{2n^5 - n+ 2000}$ are zero.
A: From your question it appears that you do have some idea of $\epsilon$ based definitions. Thus for your problem at hand, for each $\epsilon > 0$ we need to find a positive integer $N$ such that $$\left|\frac{4n^{5} - 1000}{2n^{5} - n^{3} + 2000} - 2\right| < \epsilon$$ whenever $n \geq N$. Thus we have to achieve the goal $$\left|\frac{2n^{3} - 5000}{2n^{5} - n^{3} + 2000}\right| < \epsilon\tag{1}$$ and we note that if $n > 20$ then the denominator becomes positive and we can remove absolute value signs so that the goal is changed to $$\frac{2n^{3} - 5000}{2n^{5} - n^{3} + 2000} < \epsilon\tag{2}$$ Note further that $$\frac{2n^{3} - 5000}{2n^{5} - n^{3} + 2000} < \frac{2n^{3}}{2n^{5} - n^{3}} = \frac{2}{2n^{2} - 1} < \frac{1}{n}$$ So based on this inequality our goal will be achieved if we can ensure $1/n < \epsilon$ so that $n > 1/\epsilon$. It is now clear that we can choose $N = \max(20, \lfloor 1/n\rfloor + 1)$.

Note that such problems of verifying limit definition always require you to ensure that an inequality of type $(1)$ holds when the variable $n$ is constrained in a certain way. Now most beginners try to think of the problem as solving the inequality $(1)$ via algebraic manipulation because they are so used to solving equations like $ax + b = c$ or $ax^{2} + bx + c = 0$. Note that the inequality in $(1)$ is not to be solved (i.e. find all values of $n$ for which it holds), but rather it needs to be ensured to hold by constraining $n$ in a well specified manner ($n \geq N$).
We can ensure the inequality by finding a simple expression which is greater than LHS of $(1)$ (by using trivial laws of inequalities) and then instead ensure that this newer/simpler expression is less than $\epsilon$. This will ensure that our original expression (being less than the newer expression) also remains less than $\epsilon$. Proceeding in this manner gives us the constraint on $n$ very easily and our job is done.

Another point regarding formalism. Unless you are expert in mathematical logic (or real analysis, in which case you wouldn't have asked this question) it is simply useless to write statements like these $$\forall \epsilon > 0 : \exists N : n ≥ N ⇒ |(4n^5-1000/2n^5-n^3+200) − 2| <\epsilon$$ Such formalism does not help at all in gaining any understanding or in solving such problems. Analytical arguments are much more than symbol shunting and unless one grasps the meaning of such formalism very clearly it is best to avoid it.
