2
$\begingroup$

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...

$\endgroup$
  • 3
    $\begingroup$ Divide top and bottom by $n^4$ in the original expression, then let $n \to \infty$. $\endgroup$ – астон вілла тереса лисбон Jan 25 '17 at 1:22
  • $\begingroup$ @астонвіллаолофмэллбэрг Like minds think alike? $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:24
  • 1
    $\begingroup$ @SimplyBeautifulArt think so: only their internet speeds may differ. $\endgroup$ – астон вілла тереса лисбон Jan 25 '17 at 1:26
  • 1
    $\begingroup$ @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 $\epsilon$. $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:27
  • $\begingroup$ Oh, I see. Then you have to use a bunch of inequalities to simplify things. $\endgroup$ – астон вілла тереса лисбон Jan 25 '17 at 1:30
5
$\begingroup$

$$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\Leftrightarrow \left|\frac{n^3+3n}{2n^4-n+1}\right|<\epsilon.$$ 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 $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\text{ if }n\ge \left\lfloor \frac{4}{\epsilon} \right\rfloor+1.$$

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Lol, as I am definitely too lazy at this hour in the day (US time) $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:34
  • $\begingroup$ @dxiv wow, is it that simple? Does that complete the answer or do you have to use the epsilon to show n>1/2e? $\endgroup$ – Tony Jan 25 '17 at 1:34
  • $\begingroup$ @Tony XD Well, you have to see it first. Then it's simple $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 1:35
  • $\begingroup$ @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. $\endgroup$ – dxiv Jan 25 '17 at 1:38
  • $\begingroup$ @SimplyBeautifulArt Math is where laziness is often an asset ;-) $\endgroup$ – dxiv Jan 25 '17 at 1:41
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.