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!
 A: 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.
A: are you asking for the limit $\lim_{y \to x^2} \ln (y-x^2)$?  if so, an $\epsilon-\delta$ proof should get you the contradiction you need.  (i.e., show that there exists an $\epsilon > 0$ such that for all $\delta > 0$, when $|y-x^2| < \delta$, $|f(x,y)-f(x,x^2)| \geq \epsilon$).
A: Let $\;A:=(a,a^2)\;$ be any point of $\;y=x^2\;$ (for simplicity, $\;a\neq0\;$. You can later do this particular case by yourself), and let us choose the following line
$$y=ax\;,\;\;\text{so that}\;\;\lim_{x\to a^+} (x,ax)=(a,a^2)\;,\;\;\text{then}$$
$$\lim_{x\to a^+}f(x,ax)=\lim_{x\to a^+}\log(ax-x^2)=\lim_{x\to a^+}\left(\log x+\log(a-x)\right)$$
But as the first limit on the right exists finitely whereas the second one doesn't, the left hand limit doesn't exist finitely either, and this shows the  wanted limit cannot exist (why?).
