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How to prove that the function $f(x,y)=x^y$ ($x>0$) is not continuous at the point $(0,0)$?

I tried $y=x^\alpha$ $(\alpha>0)$ but this does not work since for such $y$ we have $x^y \to 1$ as $x\to 0$.

Thank you in advance!

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Get to $(0,0)$ moving along $(x,0)$ and $(0,y)$. What happens? – Andrea Mori Mar 5 '14 at 20:12
I'm sorry, $x>0$. – Mountain Dew Mar 5 '14 at 20:16
Maybe your notation is confusing me. If you are limiting the domain of the function to x>0, then it is entirely undefined at (0,0) regardless of what its limit approaching (0,0) is. – Sparr Mar 5 '14 at 20:25
You've basically put your finger on the reason it can't be made continuous at $(0,0)$. The limit depends on the path of approach, so it doesn't exist. Good for you! – MPW Mar 5 '14 at 21:03


$$y={-a\over\ln x}\qquad a\gt0$$

Then $y\gt0$ if $x\lt1$ and $y\to0$ as $x\to0$, but $x^y=e^{-a}$.

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It is continous iff $$\lim_{(0,0)} y\log x$$ exists, but

$$\lim_{x\to 0}\lim_{y\to 0} y\log x = 0\\ \lim_{y\to 0^+}\lim_{x\to 0} y\log x = -\infty \\ $$ Hence, the limit does not exist.

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Thank you very much!!! – Mountain Dew Mar 5 '14 at 20:22
That's essentially right, but not completely right. It's continuous at $(0,0)$ if the limit exists and equals $f(0,0)$. Since the limit doesn't exist, the function can't be continuous at $(0,0)$ no matter how you define $f(0,0)$. But if the limit did exist, you still would have to define $f(0,0)$ to have continuity there (obviously, you would want to define $f(0,0) = \lim_{(x,y)\rightarrow(0,0)}f(x,y)$ for this to be the case). – MPW Mar 5 '14 at 21:01
if the limit exists, then it is $f(0,0)$, because as the limit of every sequence converging to $(0,0)$, $f(0,0) = \lim f(x_n,y_n)$ with $(x_n,y_n) = (0,0)\to (0,0)$. – mookid Mar 5 '14 at 21:05

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