- #1

- 50

- 0

## Homework Statement

Prove that the following sequence has the following limit:

[tex]\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0 [/tex]

## Homework Equations

N/A

## The Attempt at a Solution

First I did the following:

[tex]{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}[/tex]

Then:

[tex]{\frac{1}{n+3} < \epsilon[/tex]

[tex]{{n+3} > \frac{1}{\epsilon}[/tex]

[tex]{{n} > \frac{1}{\epsilon}-3[/tex]

Putting it all together:

Let [tex]\epsilon > 0[/tex] be given

Let [tex]{{n} > \frac{1}{\epsilon}-3[/tex]

Then: [tex]{\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 }[/tex]

Therefore [tex]\left|\frac{n}{n^2+3n+1}-0\right| < \epsilon [/tex]

QED

Am I going about this the right way?

Last edited: