# find $\alpha$ such that $f(x)=x^\alpha*sin(\frac{1}{x})$ is n th order derivable at (0,0)

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

$f(x,y)=(x^2+y^2)^\alpha*sin(\frac{1}{sqrt(x^2+y^2)})$

I wonder are they almost the same? Or there might be some subtle difference?

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