Prove $f(x,y) = \frac{x^2+y^2}{x+y}$ is not continuous at $(0,0)$. 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)?
 A: This answer uses a similar technique to the previous answer, but it avoids the Taylor expansion.  Consider the function $y=x^2-x$ and consider approaching $(0,0)$ along the curve defined by this function (in other words, as $x$ approaches $0$).  Then we have,
$\lim_{x\rightarrow 0}\frac{x^2+(x^2-x)^2}{x+(x^2-x)}$.
Simplifying and expanding, this becomes 
$\lim_{x\rightarrow 0}\frac{x^4-2x^3+2x^2}{x^2}$.
Simplifying, by dividing by $x^2$ gives that this limit is 
$\lim_{x\rightarrow 0}x^2-2x+2$.
Therefore, the limit along this path is 2.  Since the function is identically $0$ along the path $y=-x$, the limit along that path is $0$.  This is a contradiction because if the limit were to exist, then its value along all paths would be the same.
A: If you consider the curve given by $y=-\sin(x)$, you should run into a problem as $x$ approaches $0$.  More precisely consider
$\lim_{x\rightarrow 0}\frac{x^2+\sin^2(x)}{x-\sin(x)}$.
If we expand the Taylor series (centered at 0) for both the numerator and denominator, the numerator is $2x^2+O(x^4)$, i.e., the leading term is $2x^2$ and the remaining terms have higher powers of $x$.  On the other hand, the denominator is $\frac{x^3}{6}+O(x^5)$.  Therefore, the limit is
$\lim_{x\rightarrow 0}\frac{2x^2+O(x^4)}{\frac{x^3}{6}+O(x^5)}$.
By dividing through by $x^2/6$, we have that this limit is the same as 
$\lim_{x\rightarrow 0}\frac{12+O(x^2)}{x+O(x^3)}$.
As $x$ approaches $0$, the numerator approaches 12 while the denominator approaches 0.  Therefore, the limit does not exist.
I was able to find this example by considering the $\epsilon-\delta$ definition of a limit and noticing that a curve which approaches the $x=-y$ line as $x$ approaches $0$ would cause problems in that definition.
The other reason to develop this solution is that the numerator vanishes to second order because of the squared terms (but no higher order, if you look at the Taylor expansion).  On the other hand, the denominator has been chosen so that $x$ and $\sin(x)$ both vanish to first order and their difference vanishes to second order.
A: Hint: Since $$\forall (x,y)\in \mathbb R^2\left(x\neq -y\implies \dfrac{x^2+y^2}{x+y}=y-x+\dfrac{2x^2}{x+y}\right)$$ and $$\lim \limits_{(x,y)\to (0,0)}(y-x)=0,$$ if $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{x^2+y^2}{x+y}\right)$ existed, so would $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2}{x+y}\right)$ exist.
Now try to come up with different sublimits for $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2}{x+y}\right)$. (You can get inspiration for this from here).
A: 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 $(0,0)$.
