proving limits exist

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 $$\frac{n^3}2≥6$$.

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

5 Answers

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.

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
$$\frac{3}{n} < \epsilon \implies \frac{3}{\epsilon} < n.$$

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

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

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

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

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.