# continuity and limit in one coordinate implies limit both coordinates

I have a function that is continuous everywhere in its domain of definition, which is the plane with an axis removed. Formally,

Let $f: (\mathbb{R} \setminus \{0\}) \times \mathbb{R} \to \mathbb{R}$ be a function such that $f$ is continuous at each point $(x,y) \in (\mathbb{R}\setminus \{0\}) \times \mathbb{R}$

I know the limits of this function as we approach the y-axis, along horizontal lines:

For each $y \in \mathbb{R}$, the limit $\lim_{x\to 0}f(x,y)$ exists and $\lim_{x\to 0}f(x,y) = 0$

I want to know now does this limit property along with the continuity mean that the following also hold: $\lim_{(x,y) \to (0,0)}f(x,y) = 0$?

The difficulty seems to be that the modulus of continuity as one moves from one horizontal line to the next, towards the x-axis, could be very badly behaved. But then I think that my joint continuity should prevent this. I am not sure how to use this, however. I first thought to use some compactness argument, but I think the fact that a point is removed from the space is tripping me up.

Thanks!

• If you are taking limits as $x \rightarrow 0$, then aren't you approaching the $y$-axis along horizontal lines? – Shagnik Aug 24 '16 at 15:19

You cannot conclude that the limit exists, even when the function is defined over $\mathbb {R}^2 \setminus \{(0,0)\}$. For example, take $f (x,y) = \frac {xy}{x^2+y^2}$. This is continuous on its domain, and for fixed $y$ this converges to $0$ as $x \rightarrow 0$.
However, if we approach the origin along the curve $x=cy$ for any $c \in \mathbb {R}$, the limit is $\frac {c}{c^2 +1}$, which can take a wide range of values. Hence the limit at the origin does not exist.