epsilon -N proof of a limit help

[shelly89]

really struggling with this question...


Determine the limit of this sequences, weather it converges or diverges and prove your limit is the correct one.


\(\displaystyle a_{n} = sin(\frac{2\pi}{n}) \)

the lim as n approaches infinity is 0

so \(\displaystyle | sin(\frac{2\pi}{n})-0 | <\epsilon \)

\(\displaystyle sin(\frac{2\pi}{n}) < \epsilon \) if and only in n>N

Am not sure how to go from here...

 

[pka]

a_{n} = sin(\frac{2\pi}{n}) [/tex] the lim as n approaches infinity is 0
so \(\displaystyle | sin(\frac{2\pi}{n})-0 | <\epsilon \).

Using the mean value theorem it is easy to see that if \(\displaystyle 0<x<1 \) then \(\displaystyle \sin(x)<x \)

There is an \(\displaystyle N\in\mathbb{Z}^+ \) such that \(\displaystyle \dfrac{2\pi}{N}<\min\{1, \epsilon\} \) then WHAT?

 

[shelly89]

Using the mean value theorem it is easy to see that if \(\displaystyle 0<x<1 \) then \(\displaystyle \sin(x)<x \)

There is an \(\displaystyle N\in\mathbb{Z}^+ \) such that \(\displaystyle \dfrac{2\pi}{N}<\min\{1, \epsilon\} \) then WHAT?

where did you get \(\displaystyle \frac{2\pi}{N} \)

 

[pka]

where did you get \(\displaystyle \frac{2\pi}{N} \)

Do you know about the real numbers having the Archimedean property?

If \(\displaystyle r>0 \) then \(\displaystyle (\exists N\in\mathbb{Z}^+) \) such that \(\displaystyle \left(\dfrac{2\pi}{N}\right)<r \).

That is where I got it.