# What paths to choose to prove that $\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$ does not exist?

It's a quite simple question. But I couldn't see it...
How to prove that $$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$$ doesn't exist?
It's sufficient to show that for different paths through the origin this limit has diferent values (or doesn't exist in some of them). But, what paths to choose?!

Note: The original (and accepted) version of this answer was incorrect, owing to a slip in mental algebra on my part.

Try $y=x\sqrt{x+1}$ and $y=0$.

• Isn't the limit along both paths equal to 0 as $(x,y)\to(0,0)$? – user84413 Apr 4 '15 at 0:12
• @user84413: Yep; mental algebraic error on my part. I’ve changed it to a path that actually does work. – Brian M. Scott Apr 4 '15 at 14:02
• Thanks for your reply, and for making this change. (I think the path $y=\frac{x}{\sqrt{x+1}}$ will also work.) – user84413 Apr 4 '15 at 15:43

If we consider the path: $y^2=x^2+\left |\frac{x^3}{n-x}\right|$ for any real $n \neq 0$ we have : $$\frac{xy^2}{x^2-y^2}=\frac{x(x^2+\frac{x^3}{n-x})}{\frac{x^2}{n-x}}=\frac{x^2(n-x)+x^3}{x^2}=n$$

and here we are done because our pass through to $(0,0)$

Consider $(r,\theta)$, such that $x = r\cos(\theta), y =r\sin(\theta)$.

Then $$L(x,y) = \lim_{(x,y)\to 0}\frac{xy^2}{x^2 - y^2} = \lim_{r \to 0} \frac{r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta) -\sin^2(\theta))} = \lim_{r \to 0} r\cdot \frac{\cos(\theta)\sin^2(\theta)}{\cos(2\theta)} = L(r,\theta)$$

So clearly, even though $r\to 0$, the limit is dependant on $\theta$ since $L(r,\theta)$ is undefined for $\theta = \frac\pi4 + \frac{n\pi}2, \ n\in\mathbb{Z}$

Generally, you can see that a limit does not exists by choosing differents paths approaching to $(0,0)$.

Intuitively, imagine your $f(x,y)$ plotted. Then, those paths are functions of one variable, let's say $y=x^2$. Imagine that function plotted on the plane. then, of you do $f(x,x^2)$, you will have the curve $y=x^2$ "projected" on your $f(x,y)$.

If you approach to $(0,0)$ by different paths (for example, $y=x$, or more generally $y=mx$ (you'll have to look respectively $f(x,x)$ and $f(x,mx)$)) and you find out that the limit approaching $(0,0)$ by those paths are different, then you can say that the limit doesn't exist.

Note that you can't say that a limit exists by looking different paths, even if you've looked $100$ and you seen that the limit is equal in all the cases.

As Brian said, you can choose $y=\frac12\sqrt{x}$ and $y=0$, for example, but those are not the only paths you can choose.

There's another option, the option that used jameselmore. It consists on converting your cartesian coordinates to polar coordinates. If you see that the limit depends on the angle $\theta$, then your limit doesn't exists.

• Sure, @Abrahamlure, but WHAT paths to choose? – Ders Apr 3 '15 at 20:59
• @AndersonFelipeViveiros I've edited the answer. – Relure Apr 3 '15 at 21:08
• You're welcome @AndersonFelipeViveiros. – Relure Apr 3 '15 at 21:11

In this case just note that the function does not exist anywhere on the line $x=y$, so there cannot be a limit. Following either axis the value is $0$

• There can be a limit, since the origin is in the closure of the natural domain of $f$. – Brian M. Scott Apr 3 '15 at 20:52
• @BrianM.Scott: but there are points in the neighborhood of the origin where the function does not exist as well. – Ross Millikan Apr 3 '15 at 21:10
• True, but the notion of the limit of $f$ at the origin is still meaningful. – Brian M. Scott Apr 3 '15 at 21:11
• I thought I had to be able to approach the origin along any series of points that converge to the origin and have the function values converge to the limit. This would require $f$ to be defined in a punctured neighborhood of the origin. Along this line there aren't values to have converge. I suppose we could have an $f$ with removable singularities along some line, like $\frac {(x^2+y^2)(y-x)}{y-x}$ and say it has a limit at the origin. Here I could note that on opposite sides of $y=x$ the value is large and of opposite sign. – Ross Millikan Apr 3 '15 at 21:18
• Just define the limit to be $a$ iff for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x,y)-a|<\epsilon$ whenever $0<\sqrt{x^2+y^2}<\delta$ and $\langle x,y\rangle\in\operatorname{dom}f$. In $\Bbb R$ you might have a function defined only on the middle-thirds Cantor set minus an endpoint, and it would still be meaningful to ask whether it had a limit at the missing endpoint as a function on the Cantor set. – Brian M. Scott Apr 3 '15 at 21:23