# Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$

How to prove, using the definition of limit of a sequence, that:

$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$

Subtracting 3 and taking the absolute value of the function I have:

$$<\frac{n^3+3n}{2n^4-n}$$

But it's hard to get forward...

• Divide top and bottom by $n^4$ in the original expression, then let $n \to \infty$. – астон вілла тереса лисбон Jan 25 '17 at 1:22
• @астонвіллаолофмэллбэрг Like minds think alike? – Simply Beautiful Art Jan 25 '17 at 1:24
• @SimplyBeautifulArt think so: only their internet speeds may differ. – астон вілла тереса лисбон Jan 25 '17 at 1:26
• @астонвіллаолофмэллбэрг Well, I mean I had to type out the fractions and what-not. But according to the first line "definition of limit of a sequence", we are probably supposed to use $\epsilon$. – Simply Beautiful Art Jan 25 '17 at 1:27
• Oh, I see. Then you have to use a bunch of inequalities to simplify things. – астон вілла тереса лисбон Jan 25 '17 at 1:30

$$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\Leftrightarrow \left|\frac{n^3+3n}{2n^4-n+1}\right|<\epsilon.$$ Now, $4n^3\ge n^3+3n$ and $2n^4-n+1\ge n^4$ for all $n$ positive integer. So $$\left|\frac{n^3+3n}{2n^4-n+1}\right|\le \frac{4n^3}{n^4}=\frac{4}{n},$$ and choosing $n_0=\lfloor 4/\epsilon \rfloor+1$ we have $\dfrac{4}{n}<\epsilon$ if $n\ge n_0,$ as a consequence $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\text{ if }n\ge \left\lfloor \frac{4}{\epsilon} \right\rfloor+1.$$

Subtracting 3 and taking the absolute value of the function I have:

$$<\frac{n^3+3n}{2n^4-n}$$

Then, since $n^2 \gt 3$ and $2n^3-1 \gt n^3$ for $n \gt 1\,$:

$$\require{cancel} \frac{n^3+3n}{2n^4-n} = \frac{\cancel{n}(n^2+3)}{\cancel{n}(2n^3-1)} \lt \frac{n^2+n^2}{n^3} = \frac{2}{n}$$

You should hopefully be able to take it from there.

• Lol, as I am definitely too lazy at this hour in the day (US time) – Simply Beautiful Art Jan 25 '17 at 1:34
• @dxiv wow, is it that simple? Does that complete the answer or do you have to use the epsilon to show n>1/2e? – Tony Jan 25 '17 at 1:34
• @Tony XD Well, you have to see it first. Then it's simple – Simply Beautiful Art Jan 25 '17 at 1:35
• @Tony If you know (or can otherwise use) the squeeze theorem for sequences then you are essentially done. If you do need to write the full epsilon-delta prrof, then at least you have an easy to estimate $1 / 2n$ upper bound. – dxiv Jan 25 '17 at 1:38
• @SimplyBeautifulArt Math is where laziness is often an asset ;-) – dxiv Jan 25 '17 at 1:41

Let $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\epsilon}{4}$. If $n\geq N$, then we get \begin{align}\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3 \right|& =\left|\frac{n^3+3n}{2n^4-n+1}\right|\\&=\frac{n^3+3n}{2n^4-n+1}\\ &<\frac{n^3+3n}{2n^4-n}\\ &\leq\frac{n^3+3n}{2n^4-n^4}\\ &\leq\frac{n^3+3n^3}{n^4}\\ &=\frac{4}{n}\leq\frac{4}{N}<\epsilon. \end{align} Hope it helps.