Rational Multivariable limit

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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 $\lim_{x,y\to 0,0}$ ?

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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

$$\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{x+y}\,\,\text{does not exist}$$