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.

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- September 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. - September 12th 2009, 05:54 PMProve It
- September 12th 2009, 06:00 PMmr fantastic
- September 12th 2009, 06:28 PMcgiulz
Thanks for the prompt responses.

Is my method invalid? Or, just unorthodox?

Thanks again (Happy) - September 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.

- September 12th 2009, 06:39 PMmr fantastic
- September 13th 2009, 07:32 AMxalk
- September 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*

.

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

The statement means that for each , there exists a [tex]M>0[/tex] such that whenever .

In your case, it is not difficult to see that for**any**a number can be found such that

whenever

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

For the proof of , just take values of which are**less than**some number . - September 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! - September 14th 2009, 05:26 AMHallsofIvy
- September 14th 2009, 05:27 AMHallsofIvy
- September 14th 2009, 05:34 AMmr fantastic