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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.

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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.

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  • $\begingroup$ 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
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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.

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

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