# Proving a sequence has the given limit.

## Homework Statement

Prove that the following sequence has the following limit:

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

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

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

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

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

Last edited:

Any help would be appreciated. Thanks. :)

HallsofIvy
Homework Helper

## Homework Statement

Prove that the following sequence has the following limit:

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

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

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