Is a two-part function differentiable if the limit of one part equals the value at the other part?

Question

Suppose you have a function like:

1. $f(x,y) = g(x,y)$ IF $(x,y) != (0,0)$, and

2. $f(x,y) = 0$ IF $(x,y) = (0,0)$

Can I say $f(x,y)$ is differentiable at $(0,0)$ if the limit of $g(x,y)$ as $(x,y)$ approaches $(0,0)$ is equal to $0$? i.e. $$\lim_{(x,y)\to(0,0)} g(x, y) = 0$$ This makes sense in my mind, seeing as that would make the function continuous at the point and thus differentiable, but is that correct mathematically?

Answer 1

You can say that the function is continuous at $(0,0)$, but a function can be continuous and not differentiable. ( think at a cone with vertex in the origin: $z=a^2\sqrt{x^2+y^2}$).

Answer 2

Not necessarily. As a counterexample, consider the function $$y = |x|$$ $y$ is differentiable at $x = 0^-$ and at $x = 0^+$. However, note that $y = 0$ at $x = 0$, but it is not differentiable.

Answer 3

Note that differentiability is stronger than continuity.

If a function is differentiable at a point then it is continuous at that point.

On the other hand if a function is continuous at a point it may or may not be differentiable.

A good example is $f(x) =|x|$ which is continuous at $x=0$ but it is not differentiable at $x=0$