Question

Real Analysis: Prove that the limit of sin(n)/n=0

Answer 1

So far, I have the following:
Wms that \[\forall \epsilon >0, \exists \mathbb{N} \] s.t. if n>N, \[\left| \frac{ \sin(n) }{ n } -0\right|<\epsilon \]
Let \[\epsilon >0\] be given.
\[-\epsilon <\frac{ -1 }{ n }\le\frac{ \sin(n) }{ n }\le \frac{ 1 }{ n }<\epsilon \]

Answer 2

Have you gone over the Squeeze theorem?

I believe you're going to need it to finish your problem.

Answer 3

Yes.... I have toyed with it, but will it work to prove this according to the definition of a limit? I'm just not sure if that's as in depth as I need to go. I should somehow end with n (< or >) something, by manipulating the inequality that I have, I think.

Answer 4

Could I use the squeeze theorem to force epsilon to be 1?

Answer 5

Eh, I guess you don't really need it. It's just much easier, since the lim (1/n) is incredibly easy to prove is 0. But, anyway:

Notice in your proof that you did not define N. Define it in terms of epsilon.

Answer 6

(Only reason why I suggested Squeeze Theorem is because it usually comes before the precise definition of a limit. It isn't as rigorous though, which you may need. I don't know the expectations of your class.)

Answer 7

Well, we haven't been using the squeeze theorem in this course, but have in a previous. I will see what I can come up with.

Answer 8

Ignore the Squeeze Theorem comment then. Let N = 1/epsilon.

Look at your inequality. What can you say?

Answer 9

\[-\frac{ 1 }{ N }<-\frac{ 1 }{ n }\le \frac{ \sin(n) }{ n }\le \frac{ 1 }{n }<\frac{ 1 }{ N }\]

Answer 10

ohh, say \[n >\frac{ 1 }{ \epsilon } \]

Answer 11

Yes. You're trying to prove |sin(n)/n - 0| < epsilon. You don't assume it. All you can assume is that if epsilon > 0, then you can find an N s.t. n > N.

The idea is that your epsilon is getting smaller faster than your N is getting bigger as you take larger and larger values of n. (In other words, it's converging. If you look at the graph, you'll see the oscillations get smaller and smaller the farther you go on the x-axis.)

So, let epsilon > 0 be given. Then, there exists N s.t n > N. Define N = 1/epsilon. Then,

|f(n) - 0| = |sin(n)/n - 0| = |sin(n)/n| <= |1/n| = 1/n (since n>0) < 1/N (since n > N) = epsilon (since N = 1/epsilon).

You should have something similar to that. It's very important that you don't assume what you're trying to prove though. Quite a few people make this mistake (happened to me often).

Answer 12

That small detail makes this a lot easier for me.

Answer 13

Sorry if that was confusing. The definitions take a bit to get used to. The idea for any limit proof is to define the term you're given in terms of epsilon so you can write your inequality. There's usually annoying tricks which you'll pick up the more you do them.

Answer 14

I think I understand it now though. I just have to work it out backwards kind of in order to write the formal proof out.... and I agree with your annoying tricks description. Pretty soon they won't be tricks though, they will just be normal.