Two variables limit using $\varepsilon$ and $\delta$ I'm trying to solve this limit
$$\lim_{(x,y)\to(0,0)}\frac{xy}{y-x^{3}}$$
I know the answer is $0$ (Wolfram told me). I tried to see some limited function to solve it, but I couldn't.
I mean, I tried both:
$$\frac{x}{y-x^{3}}$$
and
$$\frac{y}{y-x^{3}}$$
and they are not limited.
I really don't know how to use $\varepsilon$ and $\delta$ to prove it. From James Stewart's calculus book, I know I have to try to use $\sqrt{x^{2}+y^{2}}<\delta$ to limit $|f(x,y)-0|$ and get a link between $\varepsilon$ and $\delta$, but I really didn't get it.
 A: Hint: What happens along the curve $y=x^3 + x^5$ as $x\to 0?$
A: These two-variable limits are tricky, but you can get a handle on them by looking at a contour plot.
Let $f(x,y) = \frac{xy}{y-x^3}$.  If $f(x,y) = k$ for some constant $k$, then
$$
 xy = ky - k x^3 \implies y = \frac{kx^3}{k-x}
$$
As $k$ varies, the curves with this equation will all intersect at $(0,0)$.  But that is a Bad Thing for a function.  It means we cannot define a value for $f$ at $(0,0)$ that would result in a continuous function.  The point $(0,0)$ would have to simultaneously be on the level curve for all constants $k$.
We can also use this to show that $\lim_{(x,y) \to (0,0)} \frac{xy}{y-x^3}$ does not exist.  Use two of the curves with $k=0$ and $k=1$ (for instance).


*

*Let $x=t$ and $y=0$.  Then
$$
\lim_{t\to 0} f(t,0) = \lim_{t\to 0} 0 = 0
$$

*Let $x=t$ and $y = \frac{t^3}{1-t}$.  Then
$$
\lim_{t\to 0} f\left(t,\frac{t^3}{1-t}\right) = \lim_{t\to 0} 1 = 1
$$
Therefore the two-variable limit does not exist.
A: Rewrite your function as
$$\frac{x}{1-\frac{x^3}{y}}$$
Now, for any $a \in \Bbb{R}$, along the curve $$1- \frac{x^3}{y} = ax$$
or equivalently along the curve
$$y=\frac{x^3}{1-ax}$$
your function is constantly $a$. By arbitrarity of $a$, the limit cannot exist.
