Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:

approaching by the y-axis:

$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$

approaching by $y=x$

$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$

So I think,that this limit exists.
is it correct in this form?
HINT:
$$
\frac{x^3+y^3}{x^2+y^2}=x\frac{x^2}{x^2+y^2}+y\frac{y^2}{x^2+y^2}
$$

But your method doesn't answer the question.
Try coming at it from any other direction, e.g. y = mx.

Ced