ok just reiterating the def'n of continuity epsilon/delta style. f is cont @ a if for an epsilon > 0 there exists a delta > 0 s.t. |x-a|< delta implies |f(x) - f(a)| < epsilon.

Question:
f(x) = sin (1/x), x not = 0. f(0) = 0

need to prove that it't no continuous at 0 by finding a specific epsilon that has no reply.

so far I worked it down to this

|f(x) - f(a)| < epsilon
|f(x) - f(0)|
|f(x)|
|sin (1/x)| < epsilon

|x-a|< delta
|x-0|
|x| < delta

now I am assuming I need to choose my epsilon based on delta, ie, choosing epsilon = delta/3 or delta/6. Question is, which one do I pick and how do I verify it?

Thanks in advance

Do not overwork yourself. If n is odd

yah I mean as you look at the graph it oscillates between 1 and -1 frequently, but making a graphical arguement doesn't really work, didn't realize that it was easy enough to just choose and Epsilon < 1 would work.