Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) \over (4n^7 − 7n^2 + 5)} $ = 1/2

Question

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.

Answer 1

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}

Answer 2

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 $$\lim_{n\to\infty}\Biggl(\frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}\Biggr)=\lim_{n\to\infty}\Biggl(\frac{2+3\frac{1}{n^2}+4\frac{1}{n^6}}{4-7\frac{1}{n^5}+5\frac{1}{n^7}}\Biggr)=\frac{2+0+0}{4-0+0}=\frac{1}{2}$$