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

  • Thread starter Harrisonized
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In summary: There is no contradiction. The limit of x*sin(1/x) at x=0 does not equal 1, although the series does diverge as x shrinks to 0.
  • #1
Harrisonized
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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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  • #2
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
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
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.
 
  • #5
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?
 
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FAQ: Incorrect proof for lim sin(1/x) at x=0

What is the incorrect proof for lim sin(1/x) at x=0?

The incorrect proof for lim sin(1/x) at x=0 is that it states that the limit is equal to 1, when in fact, the limit does not exist at x=0.

Why is the proof for lim sin(1/x) at x=0 incorrect?

The proof is incorrect because it does not take into account the oscillating behavior of the function near x=0, which causes the limit to not exist.

How is the correct proof for lim sin(1/x) at x=0 different from the incorrect one?

The correct proof takes into account the oscillating behavior of the function and shows that the limit does not exist at x=0, rather than incorrectly stating that it is equal to 1.

Can the incorrect proof for lim sin(1/x) at x=0 still be used to approximate the limit?

No, the incorrect proof cannot be used to approximate the limit as it provides a wrong value and does not take into account the true behavior of the function.

Why is it important to have a correct proof for lim sin(1/x) at x=0?

Having a correct proof is important because it ensures that we have a correct understanding of the behavior of the function at x=0. It also allows us to accurately approximate the limit and make correct conclusions about the function.

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