3
$\begingroup$

Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator. For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.

it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$. this is where I have a problem. I don't understand how we got to the point where all we need is $\frac{n^3}2≥6$.

$\endgroup$
1
  • 1
    $\begingroup$ Do you mean $\lim\limits_{n\to\infty}\frac{4n^3+3n}{n^3-6}$? $\endgroup$ – robjohn Jan 17 at 11:53
1
$\begingroup$

If $\frac{n^3}2\ge 6$ then $$n^3-6\ge n^3-\frac{n^3}2=\frac{n^3}2 $$ is your desired monomial lower bound for the denominator.


In fact, we are allowed to be wasteful and may begin right away with, say, assuming $n>1000$. That makes $3n+24<3.024 n$ and $n^3-6>0.999999994n^3$ and so the error term certainly $<\frac{3.025}{n^2}$. But that doesn't matter for the sole purpose of proving the limit where any other bound of the form $<\frac C{n^2}$ (or even weaker) is good enough.

$\endgroup$
1
$\begingroup$

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.

For $n > 9$, $0 < (n^3 - 64n),$ which is a lower bound for the denominator.
For $n > 9$, $(n-8) > 1 \implies$ that the numerator is
less than $3(n+8)(n-8) = 3(n^2 - 64).$

Therefore, the fraction is less than

$$\frac{3(n^2 - 64)}{n^3 - 64n} = \frac{3}{n}.$$

Therefore, choose $n$ such that $n > 9$ and
$\frac{3}{n} < \epsilon \implies \frac{3}{\epsilon} < n.$

$\endgroup$
0
$\begingroup$

Just look at the highest power of $n$ upside and downside (and divide everything by that power of $n$):

$$\lim_{n\to\infty}\frac{4n^3+3n}{n^3-6}= \lim_{n\to\infty}\frac{4+\frac{3}{n^2}}{1-\frac{6}{n^3}}=\frac{4+\frac{3}{\infty}}{1-\frac{6}{\infty}}=\frac{4+0}{1-0}=\boxed{4}$$

$\endgroup$
0
$\begingroup$

The most dominant term in numerator is $4n^3$ and the most dominant term in denominator is $n^3$ so the limit is $4$.

$\endgroup$
4
  • $\begingroup$ Isn't the most dominant term in denominator $n^3\,$? $\endgroup$ – J. W. Tanner Jan 17 at 12:48
  • $\begingroup$ Oh! sorry for the typo, I have corrected it npw. $\endgroup$ – Z Ahmed Jan 17 at 16:16
  • $\begingroup$ Did you mean $n^3$ where you typed $n*3\,?$ $\endgroup$ – J. W. Tanner Jan 17 at 18:17
  • 1
    $\begingroup$ OH! yes, very much. $\endgroup$ – Z Ahmed Jan 18 at 4:10
0
$\begingroup$

For $n \neq 0$,$$ \frac{ 4n^3+3n}{n^3-6}= \frac{ 4+\frac{3}{n^2}}{1-\frac{6}{n^3}}.$$

You should be using the properties of the limits of sequences, like here, or Rudin's PMA Theorem $3.3$. You should prove the properties first, then use them. This then reduces your epsilon-delta task significantly for questions like this one.

$\endgroup$

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.