# Incorrect proof for lim sin(1/x) at x=0

1. Aug 1, 2011

### Harrisonized

1. The problem statement, all variables and given/known data

I need to know what's wrong with the following proof:

Assume that [PLAIN]http://img16.imageshack.us/img16/4839/eq1.gif [Broken] [Broken][/URL] exists. In other words:

[PLAIN]http://img8.imageshack.us/img8/1856/eq2.gif [Broken] (1)

But:

[PLAIN]http://img801.imageshack.us/img801/8374/eq3.gif [Broken] (2)

And because sin(1/x) is an odd function:

[PLAIN]http://img24.imageshack.us/img24/8453/eq4.gif [Broken] (3)

Therefore, by (1), if [PLAIN]http://img16.imageshack.us/img16/4839/eq1.gif [Broken] [Broken][/URL] exists, then:

[PLAIN]http://img191.imageshack.us/img191/2339/eq5.gif [Broken]

[PLAIN]http://img215.imageshack.us/img215/3218/eq6.gif [Broken]

Similarly,

[PLAIN]http://img641.imageshack.us/img641/3781/eq7.gif [Broken]

[PLAIN]http://img696.imageshack.us/img696/7108/eq8.gif [Broken]

If [PLAIN]http://img16.imageshack.us/img16/4839/eq1.gif [Broken] [Broken][/URL] exists, the only value at which the limit can exist is 0. Since the limit converges to a single value, the limit exists and is equal to 0.

2. Relevant equations

[PLAIN]http://img812.imageshack.us/img812/6119/eq10.gif [Broken]

3. The attempt at a solution

The Laurent series disagrees. >:(

I know there's something wrong with the proof, since it's well accepted that the limit doesn't exist. I'm just not sure what. Any help is appreciated.

Last edited by a moderator: May 5, 2017
2. Aug 1, 2011

### Dick

The problem is that you assume that the limit exists. It doesn't. If you assume something false to begin with, isn't it hard to trust the conclusion you would draw from that?? It is true that if an odd function has a limit at 0 then the limit must be 0, as you've shown. But sin(1/x) doesn't have a limit at 0.

3. Aug 1, 2011

### Harrisonized

The assumption was only necessary at the beginning to show that the limit, if it exists, converges to a single point. If the limit doesn't exist, shouldn't it converge to different values from different sides?

4. Aug 1, 2011

### Dick

The limit doesn't converge to a single value on either side. Look at a graph.

5. Aug 6, 2011

### Harrisonized

Thank you, Dick, for your help on the previous problem. The answer seemed obvious after I switched sin(1/x) into f(x) and reconstructed the proof for f(x).

I didn't want to make a new thread for such a related question, so here goes...

The limit of x*sin(1/x) at x=0 is 0 by the squeeze theorem. The series expansion, however, is:

x*(x-1-x-3/3!+x-5/5!-x-7/7!+... )

= 1-x-2/3!+x-4/5!-x-6/7!+...

Is there a contradiction? The series seems to diverge as x shrinks to 0. If the limit of x*sin(1/x) is truly 0, then the limit of

x-2/3!-x-4/5!+x-6/7!-...

must equal 1 when x=0. Is there a way to show this?

Last edited: Aug 6, 2011
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