TITLE: Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$
QUESTION [6 upvotes]: 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

REPLY [2 votes]: Let $y=(x^3-x^6)^2-x^2$. $\,$Then
\begin{align*}
\frac{x^3+y^3}{x^2+y} &= \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}\\
&= \frac{1}{x^3-x^6} + (x^3-x^6)^4+3x^4-3x^2(x^3-x^6)^2\\
&\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.