Assume we have a function $f: \mathbb R^2\to \mathbb R$, such that $f(t,0) = f(0,t) = 0$ for all $t\in \mathbb R$ and $f(x,y) = 1$ for $x \neq 0$ and $y\neq 0$.
I was told that the function is continuous at point $(0,0)$, but I can't get it. As far as I understand, the function is only continuous if you can imagine a circle around the point $(0,0)$ and that the limit - no matter from which direction you approach - exists and equals to the value at $(0,0)$. Since the function is of value $1$ outside the "cross" (thats how I imagine the function), there is a discontinuity, right?
Does it make any difference if we put $f(t,0) = f(0,t) = 1$ and for all others points $f(x,y) = 0$? Another example would be $f(t,0) = f(0,t) = t$ and for all other points $f(x,y) = 1$?
The only example I can think of where the function is continuous at point $(0,0)$ is the following example: $f(t,0) = f(0,t) = t$ and $f(x,y) = 0$ for all other points in the $\mathbb R^2$ plane. Is that correct?
But in all above mentioned examples the function would be partially differentiable at point $(x,y) = (0,0)$, right?