TITLE: Prove $f(x,y) = \frac{x^2+y^2}{x+y}$ is not continuous at $(0,0)$.
QUESTION [2 upvotes]: Let $f(x,y) = \frac{x^2+y^2}{x+y}$ when $x+y \neq 0$ and $f(x,y) = 0$ when $x+y=0$. Prove $f$ is not continuous at $(0,0)$ in the $R^2$ norm.
Is this as easy as noticing that the function is undefined on the line $y=-x$ and hence
$$\lim_{x \to 0} f(x,-x) = \frac{2x^2}{x-x}$$
does not exist? Should this not be true in any norm then (even though the question specifically asks for the case of the $R^2$ norm)?

REPLY [1 votes]: We can begin with $\lim_{x \to 0}\dfrac{x^2+y^2}{x+y}=\lim_{x \to 0}\dfrac{(x+y)^2-2xy}{x+y}$, as this allows us to focus only on $\lim_{x \to 0}\dfrac{-2xy}{x+y}$. This is obviously defined for $x+y\gt0$, and, as per the question, is also defined for $y=-x$. We therefore need to consider $y=kx$ as $k\to-1$. This allows us to write $\lim_{x \to 0,k \to -1}\dfrac{-2kx^2}{x(1+k)}=\lim_{x \to 0,k \to -1}\dfrac{-2kx}{(1+k)}$. But as $k \to -1$ for any $x\gt0$, we can make this function as large as we want, and so $f(x,y)$ is not continuous at $(0,0)$.
{"set_name": "stack_exchange", "score": 2, "question_id": 1169199}