# If $g(x):=f(x, kx^m)$ is continuous at $0$, then $f(x,y)$ is continuous at $(0,0)$

If $g(x):=f(x, kx^m)$ is continuous at $0\;\;\;$ $\forall k\in R$, $\;\;\;\forall m\in N$, then $f(x,y)$ is continuous at $(0,0)$.

I'm not quite sure what is meant here by at 0. This means $f(x, kx^m) = 0$? Seems there could be an issue in this statement when k = 0?

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What is meant by "at $0$" is that $\lim_{x\to 0}g(x)=g(0)$. That is, the hypothesis is that for all real $k$ and positive integers $m$, $\lim_{x\to 0}f(x,kx^m)=f(0,0)$, and you want to prove that $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$. – Jonas Meyer Jan 30 '11 at 2:59

It isn't true. Let $f$ be defined by $f(x,y)=\frac{xy^2}{x^2+y^4}$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$. I got this example by Googling "continuous on each line" and clicking on this. You can see that $f$ is not continuous at zero by approaching $(0,0)$ along the curve $x=y^2$, but it satisfies the hypotheses of the problem.
There's a much easier example. Define $f$ by $f(0,y)=1$ if $y\neq0$ and $f(x,y)=0$ otherwise.
To refine Jonas' answer, you could take $f(x,y)=\frac{e^{-1/x^2}y}{e^{-2/x^2}+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$. Then the limit along any algebraic curve passing through $(0,0)$ is zero, but the limit along $y=e^{-1/x^2}$ is $1/2$.