# Help with a limit

## Main Question or Discussion Point

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.

JasonRox
Homework Helper
Gold Member
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?

HallsofIvy
Homework Helper
What exactly is "Haines definition of the limit"?

JasonRox
Homework Helper
Gold Member
What exactly is "Haines definition of the limit"?
His professors name?

matt grime
Homework Helper
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

matt grime
Homework Helper
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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ej matt grime thank you, this is what i was not quite sure about. Now it is all clear thnx for your help guys