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

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Homework Help Overview

The discussion revolves around the limit of sin(1/x) as x approaches 0, with participants examining a proof that claims the limit exists and is equal to 0. The subject area is calculus, specifically the behavior of limits involving oscillatory functions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the validity of the initial assumption that the limit exists, with some noting that assuming a false premise undermines the proof's conclusions. Others explore the implications of the limit not existing and discuss the behavior of the function from both sides of the limit.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the limit and the implications of the proof's assumptions. There is recognition that the limit does not exist, and some participants are exploring related concepts, such as the limit of x*sin(1/x) at x=0 and its series expansion.

Contextual Notes

Participants are grappling with the implications of the limit's non-existence and the behavior of the function sin(1/x) as x approaches 0. There is mention of the squeeze theorem in relation to a different limit, indicating a potential area of confusion regarding series expansions and convergence.

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Homework Statement



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

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

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

But:

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

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

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

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

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

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

Similarly,

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

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

If [PLAIN]http://img16.imageshack.us/img16/4839/eq1.gif [/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.

Homework Equations



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

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.
 
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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.
 
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?
 
Harrisonized said:
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?

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

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