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Perhaps it's for all x in its domain?

the function is not continuous at x=0, are you missing something?

it's for \[f(x) \rightarrow R:R\]

that I know. that's because x while x-->0 equals 0 and sin(1/x) is a Bounded function

that's what I thought

f(x) = x sin(1/x) is particularly interesting ... being continuous but not differentiable at x=0.

which is the case that you're presenting

yeah ... the case is different from what you are asking.

let me ask this to experts!! I'll reply back later.

tnk u

http://chat.stackexchange.com/transcript/message/6829811#6829811