I don't believe this is correct. That function isn't defined at 0. While it does have a limit, continuity at a point c requires that the limit as x->c = f(c), which obviously isn't the case here.
So just define it to 0. You are correct but it's a minor detail.
Re:Calculus Is Deeply Irrational
by OscarCunningham on Wednesday August 21, 2019 @02:26AM
(#20754964)
Can a function have that property at every point?
x sin(1/x) is continuous and non-differentiable at 0, but I wouldn't say it has a corner there.
A sharp corner to me means a point where the left and right derivatives both exist but are different.