# Epsilon Limit Question

• Nov 1st 2010, 12:24 PM
craig
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
• Nov 1st 2010, 12:30 PM
TheEmptySet
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
• Nov 1st 2010, 02:49 PM
craig

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
• Nov 1st 2010, 03:42 PM
TheEmptySet
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
• Nov 2nd 2010, 11:23 AM
craig
Thankyou that does actually help, not thought about setting it as a "challange" before.