Using the definition of a limit, prove that $$\lim_{n \rightarrow \infty} \frac{n^2+3n}{n^3-3} = 0$$

I know how i should start: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$

$$\left |\frac{n^2+3n}{n^3-3} - 0 \right | < \epsilon$$

but from here how do I proceed? I feel like i have to get rid of $3n, -3$ from but clearly $$\left |\frac{n^2+3n}{n^3-3} \right | <\frac{n^2}{n^3-3}$$this is not true.


This is not so much of an answer as a general technique.

What we do in this case, is to divide top and bottom by $n^3$: $$ \dfrac{\frac{1}{n} + \frac{3}{n^2}}{1-\frac{3}{n^3}} $$ Suppose we want this to be less than a given $\epsilon>0$. We know that $\frac{1}{n}$ can be made as small as we like. First, we split this into two parts: $$ \dfrac{\frac{1}{n}}{1-\frac{3}{n^3}} + \dfrac{\frac{3}{n^2}}{1-\frac{3}{n^3}} $$

The first thing we know is that for large enough $n$, say $n>N$, $\frac{3}{n^3} < \frac{3}{n^2} < \frac{1}{n}$. We will use this fact.

Let $\delta >0$ be so small that $\frac{\delta}{1-\delta} < \frac{\epsilon}{2}$. Now, let $n$ be so large that $\frac{1}{n} < \delta$, and $n>N$.

Now, note that $\frac{3}{n^3} < \frac{3}{n^2} < \frac{1}{n} < \delta$. Furthermore, $1- \frac{3}{n^3} > 1 - \frac{3}{n^2} > 1-\delta$.

Thus, $$ \dfrac{\frac{1}{n}}{1-\frac{3}{n^3}} + \dfrac{\frac{3}{n^2}}{1-\frac{3}{n^3}} < \frac{\delta}{1+\delta} + \frac{\delta}{1+\delta} < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon $$

For large enough $n$. Hence, the limit is zero.

I could have had a shorter answer, but you see that using this technique we have reduced powers of $n$ to this one $\delta$ term, and just bounded that $\delta$ term by itself, bounding all powers of $n$ at once.

  • $\begingroup$ I am hoping that you have understood the above process. As an exercise, or just for fun, try the limit $\displaystyle\lim_{n \to \infty} \frac{n^4+7n^6+35n+42}{12n^3+34n^2+278n^6 + 1728}$. The answer is just the coefficients of the largest powers, $\frac{7}{278}$. If you can answer this, then you will be able to appreciate this technique. $\endgroup$ – астон вілла олоф мэллбэрг Sep 28 '16 at 0:40
  • $\begingroup$ yeah I think I get an idea from your proof. but my professor never proved something using $\delta$ yet. so I was wondering if you could prove this without using $\delta$ $\endgroup$ – Allie Sep 28 '16 at 0:48
  • $\begingroup$ Of course. See, $\delta$ is just a "proxy name" for $\frac{1}{n}$. In place of $\delta$, I could have just taken $\frac{1}{n}$ in that expression, and then I will directly get a bound for $\frac{1}{n}$ instead of $\delta$. Just for ease of reading, I decided to split this into two parts, one having $\delta$ and one having $\frac{1}{n}$.Hence, $\delta$ can be avoided easily, $\endgroup$ – астон вілла олоф мэллбэрг Sep 28 '16 at 1:41

Another way to show $\lim_{n \rightarrow \infty} \frac{n^2+3n}{n^3-3} = 0 $.

Let $f(n) =\frac{n^2+3n}{n^3-3} $.

For $n \ge 3$, $n^2+3n < 2n^2 $.

Similarly, for $n \ge 3$, $n^3-3 = \frac12 n^3 + \frac12 n^3 - 3 \gt \frac12 n^3 $ for $\frac12 n^3 > 3$, which is certainly true for $n \ge 2$.

Therefore, for $n \ge 3$, $f(n) =\frac{n^2+3n}{n^3-3} < \frac{2n^2}{\frac12 n^3} =\frac{4}{n} $.

You should now be able to easily show that $\lim_{n \to \infty} f(n) = 0 $.

Note that this is not the best upper bound for $f(n)$, but it is enough to show what you want.

The generalizations are left to you.


How about this: when $n>2$, $0<\frac{n^2+3n}{n^3-3}<\frac{4n^2}{n^3-n^2}=\frac{4}{n-1}$


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.