How to prove that $f(x,y)=\ln{(y-x^2)}$ has no limit at $y=x^2$ I try to find a way to prove that $f(x,y)=\ln{(y-x^2)}$ has no limit at $y=x^2$ (it's border).
Can you direct me and give my ways to show it??
Thank you!

For some fixed $x$: $$y \to x^2 \implies z \equiv y-x^2 \to 0$$
So: $$\lim_{y \to x^{2^{+}}} \ln({y-x^2}) = \lim_{z \to 0^{+}} \ln({z}) = - \infty$$
The left sided limit doesn't exist because $\ln$ is simply not defined for non-positive numbers.
