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

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In summary, In this conversation, the following was proved: 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
  • #1
sutupidmath
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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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  • #2
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?
 
  • #3
What exactly is "Haines definition of the limit"?
 
  • #4
HallsofIvy said:
What exactly is "Haines definition of the limit"?

His professors name? :confused:
 
  • #5
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.
 
  • #6
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
 
  • #7
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).
 
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  • #8
ej matt grime thank you, this is what i was not quite sure about. Now it is all clear thnx for your help guys
 

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

What is the limit of sin(1/x) as x approaches 0?

The limit of sin(1/x) as x approaches 0 does not exist. This means that no single value can be assigned to the limit as x gets closer and closer to 0.

How can you prove that the limit of sin(1/x) does not exist?

The limit of sin(1/x) can be proven to not exist by showing that the function oscillates or approaches different values as x approaches 0 from different directions. This can be done through the use of the squeeze theorem or by examining the graph of the function.

Why is it important to prove that the limit of sin(1/x) does not exist?

Proving that the limit of sin(1/x) does not exist is important in understanding the behavior of this function and its limits. It also has implications in calculus and other areas of mathematics where limits are used.

What is the difference between the limit of sin(1/x) and the limit of 1/x as x approaches 0?

The limit of sin(1/x) does not exist, while the limit of 1/x as x approaches 0 is equal to 0. This is because the function sin(1/x) oscillates and does not approach a single value as x gets closer to 0, while the function 1/x approaches 0 as x approaches 0.

Can the limit of sin(1/x) be proven to exist?

No, the limit of sin(1/x) cannot be proven to exist. This is because the function oscillates and does not approach a single value as x gets closer to 0, which is a requirement for the existence of a limit.

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