Rational Multivariable limit I am having some issues with the following multivariable limit:
$$\lim_{x,y\to0,0} \frac{x^2+y^2}{x+y}$$
I am trying to show whether it exists and is equal to 0, or whether it does not exist. What I tried to do was convert it to polar coordinates and then show the limit was zero from there, however I am not sure if what I have done is valid. Here is how I have done it:
Rewrite in terms of polar coords:
$$\lim_{r\to0^+} \frac{r^2}{r\sin(\theta) + r\cos(\theta)} = \frac{r}{\sin(\theta) + \cos(\theta)}$$
We know that for all $\theta$ that does not make the denominator $0$ (i.e everywhere in the domain of the function), this limit must be $0$. Is this correct? More specifically, is $\lim_{r\to0^+}$ replacable with $\lim_{x,y\to 0,0}$ ? 
 A: Let a curve $C$ be described parametrically as $x=t$ and $y=-t+t^3$.  Note that the curve approaches the origin as $t\to 0$.  
We have on $C$
$$\frac{x^2+y^2}{x+y}=\frac{t^2+(-t+t^3)^2}{t^3}=t^3-2t+\frac2t$$
Thus, on $C$ the limit of interest is undefined since the limit from the right is
$$\lim_{t\to 0^+}\left(t^3-2t+\frac2t\right)=\infty$$
while the limit from the left is 
$$\lim_{t\to 0^-}\left(t^3-2t+\frac2t\right)=-\infty$$
We conclude that the limit
$$\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{x+y}\,\,\text{does not exist}$$
A: If we approach the origin along the line $y=kx$ the limit is $0$ :
$$\lim_{(x,y)→(0,0)}\frac{(x^2+y^2)}{(x+y)}=\lim_{(x,kx)→(0,0)}\frac{(x^2+(kx)^2)}{(x+kx)}=\lim_{(x→0)}\frac{(x(1+k^2))}{(1+k)}=0$$
If we approach the origin along the curve $y=ax^2+bx$  we have :
$$\lim_{(x,y)→(0,0)}\frac{(x^2+y^2)}{(x+y)}=\lim_{(x,ax^2+bx)→(0,0)}\frac{(x^2+(ax^2+bx)^2)}{(x+ax^2+bx)}=
\lim_{(x→0)}\frac{x(1+b^2+2abx+a^2 x^2)}{(1+b+ax)}\to(1)$$
Now , if $b=+1$ the limit is again $0$.
If $b=-1$ expression  (1) becomes :
$$\lim_{(x→0)}\frac{x(2-2ax+a^2 x^2)}{ax}=\lim_{(x→0)}\frac{2-2ax+a^2 x^2}{a}=\frac2a$$
So if for $b=-1$ we let, say $a=1$, we can choose to approach the origin along the curve $y=  x^2-x$ and we get
$$\lim_{(x→0)}\frac{2-2ax+a^2 x^2}{a}=\lim_{(x→0)}{(2-2x+x^2)}=2$$Since two different limit values on different path approach, the limit does not exist.
