How can I show that $F(x,y)$ is not continuous? I want to show
$F(x,y)=\frac{xy}{x^2+y^2}$ is not continuous at the origin. Here is my attempt:
Choose a path $y=mx$. Then, $F(x,mx)=\dfrac{xmx}{x^2+m^2x^2}=\dfrac{mx^2}{x^2(1+m^2)}=\dfrac{m^2}{1+m^2}$ if $(x,mx)\neq0$. Thus, $\lim_\limits{(x,y)\to(0,0)}F(x,y)$ does not exist, so $F$ is not continuous.
 A: If your function was $F(x,y)=\dfrac{xy}{x^2+y^2}$, you're correct indeed, $F$ is not continuous at $(0,0)$, since we have a different limit for each path $y=mx$.
To be more specific, for each $m$ we can consider two paths, $y=mx$ and $y=-mx$, which will give the same limit, but limits are distinct for another path $y=nx$ if $|m|\neq|n|$.
Strictly speaking, to talk about the continuity of $F$ at $(0,0)$ you should have defined $F$ there, as it was pointed out in the comments. But you could reformulate the question as: Could $F$ de defined at $\mathit{(0,0)}$ so it is continuous there? And the answer would be no because of what you showed: different paths have different limits, so we can't make it continuous.
A: Hint:
Consider first the two sequences $(x_n)_{n \in \mathbb N^\times}$ and $(y_n)_{n \in \mathbb N^\times}$ defined as
$$
x_n = \frac 1 n \quad \text{and} \quad y_n = \frac 1 n \; .
$$
Note that these are both zero sequences. Now consider $F(x_n, y_n)$ and calculate
the limit
$$
\lim_{n \to \infty} F(x_n, y_n) \; .
$$
Next, consider the two sequences $(x'_n)_{n \in \mathbb N^\times}$ and $(y'_n)_{n \in \mathbb N^\times}$ defined as
$$
x'_n = \frac 1 n \quad \text{and} \quad y'_n = -\frac 1 n \; .
$$
Again, calculate the limit
$$
\lim_{n \to \infty} F(x_n', y_n') \; .
$$
Deduce from that, that the function $F$ is not continuous at $(0,0)$.
