# Prove a limit

• Sep 12th 2009, 05:35 PM
cgiulz
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 PM
Prove It
Quote:

Originally Posted by cgiulz
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?

Thanks.

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 PM
mr fantastic
Quote:

Originally Posted by cgiulz
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.

The usual thing to do is $\displaystyle \frac{n}{n^2 + 3n + 1} = \frac{1}{n + 3 + \frac{1}{n}} \rightarrow 0$ as $\displaystyle n \rightarrow \infty$.
• Sep 12th 2009, 06:28 PM
cgiulz
Thanks for the prompt responses.

Is my method invalid? Or, just unorthodox?

Thanks again (Happy)
• Sep 12th 2009, 06:39 PM
Krizalid
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 PM
mr fantastic
Quote:

Originally Posted by cgiulz
Thanks for the prompt responses.

Is my method invalid? Or, just unorthodox?

Thanks again (Happy)

To apply your method properly you would need to set up the sort of inequality required by the pinching theorem (also known as the sandwich theorem or squeeze theorem).
• Sep 13th 2009, 07:32 AM
xalk
Quote:

Originally Posted by mr fantastic
The usual thing to do is $\displaystyle \frac{n}{n^2 + 3n + 1} = \frac{1}{n + 3 + \frac{1}{n}} \rightarrow 0$ as $\displaystyle n \rightarrow \infty$.

would not be better if you divided by n^2 instead n??
• Sep 13th 2009, 08:21 AM
VonNemo19
I assume from your title, and the fact that this thread is posted in a pretty advanced sub-forum, that you wish to formally prove the statement

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

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 AM
cgiulz
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 AM
HallsofIvy
Quote:

Originally Posted by mr fantastic
To apply your method properly you would need to set up the sort of inequality required by the pinching theorem (also known as the sandwich theorem or squeeze theorem).

Since you had already proved that |n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n all you need to do is observe that, since n is a positive integer, 0< |n/(n^2 +3n + 1)| < 1/n and both 0 and 1/n go to 0 as n goes to infinity.
• Sep 14th 2009, 05:27 AM
HallsofIvy
Quote:

Originally Posted by xalk
would not be better if you divided by n^2 instead n??

Good heavens, no! then you would have both numerator and denominator going to 0 and "0/0" is as indeterminate as "$\displaystyle \infty/\infty$".
• Sep 14th 2009, 05:34 AM
mr fantastic
Quote:

Originally Posted by HallsofIvy
Good heavens, no! then you would have both numerator and denominator going to 0 and "0/0" is as indeterminate as "$\displaystyle \infty/\infty$".

(Pssst ..... Actually the denominator would go to 1 ....).