Proving Limit of sin(1/x) Does Not Exist

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Discussion Overview

The discussion revolves around proving that the limit of sin(1/x) as x approaches 0 does not exist. Participants explore this through sequences and definitions of limits, examining the behavior of sin(nπ) and sin((4n+1)π/2) as n approaches infinity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes using two sequences, x=1/nπ and x=2/(4n+1)π, to demonstrate the non-existence of the limit.
  • Another participant questions the implications of assuming the limit converges as x approaches 0, suggesting a problem arises with a specific epsilon value.
  • Several participants seek clarification on "Haines definition of the limit," indicating uncertainty about its specifics.
  • One participant asserts that the sequences converge to fixed values, specifically that sin(nπ) equals 0 and sin((4n+1)π/2) equals 1 for all n.
  • Another participant expresses concern about the validity of conclusions drawn as n approaches infinity, prompting further discussion on the nature of limits.
  • A later reply confirms that the value of sin(nπ) remains 0 regardless of n, emphasizing that the limit process does not alter fixed values.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and confusion regarding the application of limits and the sequences used. While some affirm the conclusions about the fixed values of the sine function, others remain uncertain about the implications of these results in the context of the limit.

Contextual Notes

Some participants express uncertainty about the definition of limits being used, and there is a lack of consensus on the implications of the sequences chosen for the proof.

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I have to prove that lim sin(1/x) when x->0 does not exist. i used Haines definition of the limit to prove this. I found two sequences x=1/n*pi, so when n-> infinity x->0, and the other sequence that also converges to zero that i used in this case is x=2/(4n+1)*pi, this also when n->infinity then x->0. now when we take take the corresponding sequences we get:

lim sin n*pi, n-> infinity, and lim sin (4n+1)*pi/2, as n-> infinity,

the problem is here, i am not sure at this case if i can take lim sin n*pi, n-> infinity =0 and lim sin (4n+1)*pi/2, as n-> infinity= 1, i think i should do like this. But at this case what i am courious to know is why should i take these results??

any help would do.
 
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Assume it does converge to a limit as x->0.

Now, let the epsilon be 0.1, then we should be able to find a gamma that satisfies this condition by the definition of the limit.

Can you see a problem that occurs when e=0.1?
 
What exactly is "Haines definition of the limit"?
 
HallsofIvy said:
What exactly is "Haines definition of the limit"?

His professors name? :confused:
 
You have a sequence x_n = sin npi, so that x_n=0 for all n.
You have a sequence y_n = sin (4n+1)pi/2, so that y_n=1 for all n.

They obviosuly converge, since they are constant. That x_n=sin npi is neither here nor there.
 
yeah matt grime i do understand that x_n = sin npi, so that x_n=0 for all n, and also y_n = sin (4n+1)pi/2, so that y_n=1 for all n. But my question is :

If it is still safe to jump to these conclusions when n->infinity?

thnx
 
Of course it is. Why wouldn't it be? x_n is always zero. Irrespective of what n is. Do you think that at some point sin(npi) is going to stop being zero?

Actualy scrub that. Let me put it this way. What has n going to infinity got to do with the value of sin(npi)? You appear to be thinking of things in the wrong order.

taking the limit of sin(npi) as n tends to infinity does not affect the values sin(npi) at all. How can it. They're fixed (and always 0).
 
Last edited:
ej matt grime thank you, this is what i was not quite sure about. Now it is all clear thnx for your help guys
 

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