Continuity of f(x,y)=(x^3)/(x^2 + y^2) at (0,0)

[Poopsilon]

So my text says that the piece-wise function f(x,y)=(x^3)/(x^2 + y^2) for (x,y)≠(0,0) and f(0,0)=0 is continuous due to the fact both its partial derivatives are bounded. I can't see how this follows, I mean the only point at which the reader would be worried about continuity is at the point (0,0) and to show that it is continuous at that point I would simply bound the function by g(x,y)=x and then show that this function goes to 0 as x -> 0. But why would the partial derivatives being bounded allow us to conclude that f(x,y) is continuous at (0,0)?

 

[Bacle]

I wonder if that condition is slightly weaker than the one that would guarantee
differentiability--and therefore continuity-- at (0,0), which is that the partials
exist and are continuous. AFAIK, continuous partials existing is sufficient, but
not necessary for continuity.

 

[o(1)]

I happened to stumble upon a similar theorem in one of Rudin's exercises.

Having to check continuity of the function in 0 only, I would desume it by the fact that partial derivatives are limited this way:

Take a in R^2 different from the origin, with a = (a_1, a_2)
Chose b so that b=(0,a_2).

So that the following identity:
f(0)-f(a) = ||a-b|| * (f(a)) -(f(b))/||a-b|| - |0-b||*(f(0)-f(b))/||0-b||)

When b approaches 0, given the condition on the partial derivatives, is 0 for every a.

Please do correct me if I am wrong.

/edited for misspelling