Proving a sequence has the given limit.

[The_Iceflash]

Homework Statement

Prove that the following sequence has the following limit:

\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0

Homework Equations

N/A

The Attempt at a Solution

First I did the following:

{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}

Then:
{\frac{1}{n+3} < \epsilon

{{n+3} > \frac{1}{\epsilon}

{{n} > \frac{1}{\epsilon}-3Putting it all together:

Let \epsilon > 0 be given

Let {{n} > \frac{1}{\epsilon}-3

Then: {\left|\frac{n}{n^2+3n+1}\right| - 0 = \frac{n}{n^2+3n+1} < \frac{n}{n^2+3n} = \frac{1}{n+3} < {\epsilon }

Therefore \left|\frac{n}{n^2+3n+1}-0\right| < \epsilon
QED

Am I going about this the right way?

 

[The_Iceflash]

Any help would be appreciated. Thanks. :)

 

[HallsofIvy]

Homework Statement

Prove that the following sequence has the following limit:

\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0

Homework Equations

N/A

The Attempt at a Solution

First I did the following:

{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}

Then:
{\frac{1}{n+3} < \epsilon

{{n+3} > \frac{1}{\epsilon}

{{n} > \frac{1}{\epsilon}-3

Since \epsilon-3< 1/\epsilon it would be sufficient to take n> 1/\epsilon.

Other than that, it looks good to me.

Putting it all together:

Let \epsilon > 0 be given

Let {{n} > \frac{1}{\epsilon}-3

Then: {\left|\frac{n}{n^2+3n+1}\right| - 0 = \frac{n}{n^2+3n+1} < \frac{n}{n^2+3n} = \frac{1}{n+3} < {\epsilon }

Therefore \left|\frac{n}{n^2+3n+1}-0\right| < \epsilon
QED

Am I going about this the right way?

 

[The_Iceflash]

Since \epsilon-3< 1/\epsilon it would be sufficient to take n> 1/\epsilon.

I don't follow that. How did you get \epsilon-3< 1/\epsilon and why can you take n> 1/\epsilon from it?

 

[Mark44]

I think HallsOfIvy meant 1/\epsilon - 3 < 1/\epsilon