# Thread: Epsilon Limit Question

1. ## Epsilon Limit Question

Q: Use the formal definition of convergence to prove that the sequence $\frac{\pi n + \sin{n}}{n}$ tends to a limit.

A: The formal definition is:

A sequence $x_n$ converges to a limit if, given $\epsilon > 0$, there is a natural number $N$ such that $|x_n - x| < \epsilon$.

Just not too sure on where to start with this one. I'm guessing that splitting up the fraction into $\pi + \frac{\sin{n}}{n}$ and then proving that the latter part tends to a limit would be the way to go?

Craig

2. Yes then just plug into your def, but first you need to "guess" what the limit $L$ is

$\displaystyle \bigg|\pi-\frac{\sin(n)}{n}-L \bigg|$

Then show you can make the above as small as you wish or less than epsilon

3. Thanks for the reply.

Well my educated guess will be that the limit is $\pi$, giving me $\displaystyle \bigg|\frac{\sin(n)}{n}\bigg|$

Now $\displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n}$ which is obvious.

In the solutions they've got this:

Let $\epsilon > 0$, take $N > \frac{1}{\epsilon}$, then for all $n>N$:

$\displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n} < \frac{1}{N} < \epsilon$, so the sequence tends to $\pi$.

Just wondering how they took $N > \frac{1}{\epsilon}$ in the first place?

Sorry for the confusion, really not getting this whole epsilon limit business.

Thanks again

Craig

4. This may help, think of epsilon as a challenge. Someone gives you maximal error estimate (epsilon) you need to find an N such that
$|x_n-x|< \epsilon$
is always true.

As you noted

$\displaystyle |\frac{\sin(n)}{n}| \le \frac{1}{n}$

It you want

$\frac{1}{n} \le \epsilon \implies \frac{1}{\epsilon} \le n$

Does this help

5. Thankyou that does actually help, not thought about setting it as a "challange" before.