Prove:

Limit as n --> infinity

n/(n^2+3n+1) -->0

I just don't know how to start here. What algebraic property can I use to get started?

Will this work?

|n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?

Thanks.

Printable View

- Sep 12th 2009, 05:35 PMcgiulz
Prove:

Limit as n --> infinity

n/(n^2+3n+1) -->0

I just don't know how to start here. What algebraic property can I use to get started?

Will this work?

|n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?

Thanks. - Sep 12th 2009, 05:54 PMProve It
This integral is of the indeterminate form

$\displaystyle \frac{\infty}{\infty}$

so L'Hospital's Rule can be used.

$\displaystyle \lim_{n \to \infty}\frac{n}{n^2 + 3n + 1} = \lim_{n \to \infty}\frac{\frac{d}{dx}(n)}{\frac{d}{dx}(n^2 + 3n + 1)}$

$\displaystyle = \lim_{x \to \infty}\frac{1}{2n + 3}$

$\displaystyle \to \frac{1}{\infty}$

$\displaystyle \to 0$. - Sep 12th 2009, 06:00 PMmr fantastic
- Sep 12th 2009, 06:28 PMcgiulz
Thanks for the prompt responses.

Is my method invalid? Or, just unorthodox?

Thanks again (Happy) - Sep 12th 2009, 06:39 PMKrizalid
of course it's valid too, you're just using the definition to prove that the limit is in efect, zero.

- Sep 12th 2009, 06:39 PMmr fantastic
- Sep 13th 2009, 07:32 AMxalk
- Sep 13th 2009, 08:21 AMVonNemo19
I assume from your title, and the fact that this thread is posted in a pretty advanced sub-forum, that you wish to formally

the statement*prove*

$\displaystyle \lim_{n\to\infty}\frac{n}{n^2+3n+1}$.

To do this, the epsilon-delta definition is in order.

The statement $\displaystyle \lim_{x\to\infty}f(x)=L$ means that for each $\displaystyle \epsilon>0$, there exists a [MAth]M>0[/tex] such that $\displaystyle |f(x)-L|<\epsilon$ whenever $\displaystyle x>M$.

In your case, it is not difficult to see that for**any**$\displaystyle \epsilon>0$ a number $\displaystyle M$ can be found such that

$\displaystyle |f(n)-0|<\epsilon$ whenever $\displaystyle n>M$

This implies that the limit is ,in fact, 0.

For the proof of $\displaystyle n\to{-\infty}$, just take values of $\displaystyle n$ which are**less than**some number $\displaystyle N<0$. - Sep 13th 2009, 09:00 AMcgiulz
Yes, I do and thanks. I was primarily looking for a way to get started, and I can fill in all the formalities. This is my first analysis class, so all responses are extremely useful, and somewhat ease the frustration of taking such a tedious class.

Thanks everyone! - Sep 14th 2009, 05:26 AMHallsofIvy
- Sep 14th 2009, 05:27 AMHallsofIvy
- Sep 14th 2009, 05:34 AMmr fantastic