How do I prove using the definition of limit that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$? I want to use the definition of limit in order to prove $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$, but I almost crack my head open trying to do this, it seems to be easy way but y cant find a $\delta$ such that $|\frac {x^2+y^3}{y^2 - x + xy}| \leq \epsilon$, any idea of how I might be able to find an inequality to transform the denominator in a nicer thing like $\frac{1}{\left\lVert (x,y)\right\rVert}$?
 A: The limit does not exist. You can write $$
y^{2}+xy-x=\left(  y+\frac{1}{2}x+\frac{1}{2}\sqrt{x\left(  x+4\right)
}\right)  \left(  y-\frac{1}{2}\sqrt{x\left(  x+4\right)  }+\frac{1}
{2}x\right).
$$
Take $y=x^{2}+\frac{1}{2}\sqrt{x\left(  x+4\right)  }-\frac{1}{2}x$ with $x>0$, to get
\begin{align*}
& \frac{x^{2}+\left(  x^{2}+\frac{1}{2}\sqrt{x\left(  x+4\right)  }-\frac
{1}{2}x\right)  ^{3}}{\left(  x^{2}+\sqrt{x\left(  x+4\right)  }\right)
x^{2}}\\
& =\frac{x^{2}+x^{3/2}\left(  x^{3/2}+\frac{1}{2}\sqrt{x+4}-\frac{1}{2}%
x^{1/2}\right)  ^{3}}{\left(  x^{3/2}+\sqrt{x+4}\right)  x^{5/2}}\\
& =\frac{1}{x}\frac{\left[  x^{1/2}+\left(  x^{3/2}+\frac{1}{2}\sqrt
{x+4}-\frac{1}{2}x^{1/2}\right)  ^{3}\right]  }{x^{3/2}+\sqrt{x+4}}%
\rightarrow\infty
\end{align*}
On the other hand if $y=0$ then $-\frac{x^2}{x}=-x\to 0$.
EDIT: robjohn yes there are simpler ways to show that the limit does not exist, but I was trying to show a general method. If you are computing a limit 
$$\lim_{(x,y)\to (0,0)}\frac{x^my^n}{Q(x,y)(y-f(x))^l}$$
where $Q(x,y)$ is a polynomial with $Q(x,y)>0$ for $(x,y)\ne (0,0)$ near $(0,0)$ and $f$ is a regular nonzero function with $f(0)=0$, or similar limits where the denominator vanishes along a regular curve (or union of curves) passing through $(0,0)$ then usually you can show that the limit does not exist. For example
$$\lim_{(x,y)\to (0,0)}\frac{x^{100}y^{2000}}{x(y-x)}$$
or$$\lim_{(x,y)\to (0,0)}\frac{x^{100}y^{2000}}{x(y-\sin x)}$$
A: Note that
$$
x=y^2\implies\frac{x^2+y^3}{y^2-x+xy}=\frac{y^4+y^3}{y^2-y^2+y^3}=y+1\overset{y\to0}\to1
$$
and that
$$
x=0\implies\frac{x^2+y^3}{y^2-x+xy}=\frac{y^3}{y^2}=y\overset{y\to0}{\to}0
$$
These show that the limit does not exist.
