Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit? 

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

I'm having trouble with proving that this limit really goes to $0$

thank you
Let $y=(x^3-x^6)^2-x^2$. $\,$Then
\begin{align*}
\frac{x^3+y^3}{x^2+y} &amp;= \frac{x^3+(x^3-x^6)^6 - x^6 + 3x^4(x^3-x^6)^2 - 3 x^2 (x^3-x^6)^4}{(x^3-x^6)^2}\\
&amp;= \frac{1}{x^3-x^6} + (x^3-x^6)^4+3x^4-3x^2(x^3-x^6)^2\\
&amp;\rightarrow \infty.
\end{align*}
That is,
\begin{align*}
\lim_{(x,y)\rightarrow (0, 0)} \frac{x^3+y^3}{x^2+y}
\end{align*}
does not exist.
Putting polar coordinates $x=r\sin \theta,y=r\cos \theta$
$$\lim_{r\to 0} \frac{r^3\sin^3\theta+r^3\cos^3\theta}{r^2\sin^2 \theta + r\cos \theta}=\lim_{r\to 0}\frac{r(\sin \theta + \cos \theta)}{\sin^2\theta+\frac{\cos \theta}{r}}$$
Now if $\cos \theta \not\to 0$ then $\frac{|\cos \theta|}{|r|}\to \infty$ hence the limit is $0$ if $\cos\theta \to 0$ then we have that $|\sin \theta|\to 1$ and since denominator tends to $1$ and numerator tends to $0$ again the limit is $0$ hence the limit is $0$ independent of $\theta$