<s>[INST] 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. 
 [/INST] Note that\begin{align}\left\lvert\frac{2n^7+3n^5+4n}{4n^7-7n^2+5}-\frac12\right\rvert&=\frac{\lvert6n^5+7n^2+8n-5\rvert}{\lvert8n^7-14n^2+10\rvert}\\&\leqslant\frac{26n^5}{8n^7-14n^2-10}\\&=\frac{26}{8n^2-14n^{-3}-10n^{-5}}\\&\leqslant\frac{26}{8n^2-24},\end{align}if $n>1$. Now, use the fact that\begin{align}\frac{26}{8n^2-24}<\varepsilon&\iff8n^2-24>\frac{26}\varepsilon\\&\iff n>\sqrt{\frac{13}{4\varepsilon}+3}.\end{align}
</s>