Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) \over (4n^7 − 7n^2 + 5)} $ = 1/2 Im trying to use the formal definition of a limit to prove
$\lim \limits_{x \to ∞} $$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ - ${1\over 2}$
$\left|{(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)} - {1\over 2} \right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)\over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)\over(8n^7-14n^2+10)}$ < ε
$ {(1)\over(4n^4)}$ = $ {(n^3)\over(4n^7)}$ < $ {(3n^3)\over(4n^7)}$ = $ {(6n^3)\over(8n^7)}$ < $ {(6n^3)\over(8n^7-14n^2+10)}$ < ε+7
$ {(1)\over(ε+7)}$ < $4n^4$
$ {1\over4ε+28^{1\over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1\over4ε+28^{1\over4}}$ . Then it must follow that:
$ {1\over4ε+28^{1\over4}}$ < n
$ {1\over4ε+28}$ < $n^4$
$ {1\overε+7}$ < $4n^4$
$ {1\over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with. 

I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$\frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=\frac{2+3\frac{1}{n^2}+4\frac{1}{n^6}}{4-7\frac{1}{n^5}+5\frac{1}{n^7}}$$
Note that $$\lim_{n\to\infty}\Bigl(\frac{1}{n^k}\Bigr)=0\;\forall k\in\mathbb{N_{>0}}$$
Thus $$\lim_{n\to\infty}\Biggl(\frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}\Biggr)=\lim_{n\to\infty}\Biggl(\frac{2+3\frac{1}{n^2}+4\frac{1}{n^6}}{4-7\frac{1}{n^5}+5\frac{1}{n^7}}\Biggr)=\frac{2+0+0}{4-0+0}=\frac{1}{2}$$
