assume there is a limit, L. So we have that for all there exists a such that
Consider when x>_0 and let =1/2.
Then |x|=x ->
I need a contradiction.
Thanks
Sorry, I don't quite understand. what x bar? Do you mean L. Then I don't know why f(x) must take the values + or - 1 in any (-delta,delta). Lastly you seem to imply L must be eith + or -1 meaning the distance between sin(1/x) and L will be 2 at some point. Why is this the case?
is some close to .
The proof basically states that for values infinitessimally close to , it can be said that is equal to both and , which is enough to show that you can not "zoom in" on an as you make and smaller.
You need to know that f(x) goes to L as x goes to a if and only if f(xn) goes to L for every sequence {xn} that goes to L. Is that a general theorem?
Don't know how you know 1/2npi goes to L but I will roll with it for now
sin(2npi)=0 so need delta such that if 0<|1/(2npi)|<delta then |-L|=|L|< e. But we can make e arbitarily small so it can be made less than |L| which is a contradiction. So f((1/2npi))has no limit so f(x) has no limit. Am I correct?
Given any , there exist an integer n such that . That means that every number from to are less than . For those values of x, sin(1/x) becomes and . That is, no matter how small is, there will exist values of x less than such that sin(1/x) takes on values of 1 and -1. You can't guarentee any less than 2.