Sorry for my bad English.
Define f(x)=0 when x=0;
This is a problem that our calculus teacher mentioned in the class, but now I'm not sure if I correctly understand it.
In my opinion,
as long as $\alpha$ > 0, according to squeeze theorem， The limit of f(x) at x=0 would be 0, and the function would be continuous at x=0;
As for 1th derivable, if $\alpha$ > 1, there would be lim(x->0) (f(x)-0*x)/x = 0 thus the derivative is 0 at x=0. However if x <= 1, lim(x->0) f(x)/x doesn't exist ,thus f(x) is not derivable at x=0;
And I guess if x > n (where n is integer), the function would be n th derivable.
Did I get the point or I made some mistake?
In addition, I came across a quite similar problem in multivariable calculus that is
I wonder are they almost the same? Or there might be some subtle difference?