# Showing limit does not exist using two-path test

I am new to using two-path test and my textbook only discusses it without showing any examples. I attempted to do this question below but I am not sure if I am correct. The question says to show the limit doesn't exist as $(x,y) \to (0,0)$: $$f(x,y)=\frac{xy}{|xy|}.$$

First I set $y=0$ and let $x \to 0$ and got the limit to be undefined Second I set $x=0$ and let $y \to 0$ and got the limit to be undefined

Is this how you do this test? Since limits are undefined they don't exist at this point $(0,0)$.

• It's an unfortunate example, since $f$ is undefined on the coordinate axes. Paths that also are frequently useful are those with $y = c\cdot x$ with a nonzero $c \in \mathbb{R}$. Here, choosing $c = 1$ and $c = -1$ exhibits different finite limits. – Daniel Fischer Sep 27 '13 at 14:42
• @DanielFischer I am a bit confused because I only know about the y=0 and x=0 method..I am unsure how you do it using the method you mentioned. – Raynos Sep 27 '13 at 14:45

In your argument, you are choosing points which are not in the domain of the function. This is illegal. What you want to do is approach $(0,0)$ via points that lie in the domain. For instance, if you choose a line $y=mx$, then along this line, $$f(x,y) = \frac{m}{|m|}$$ for any point $x\neq 0$. Thus, the limit value will depend on the sign of $m$!

So consider the line $y=x$, and you will end up with $$f(x,y) = 1$$ for all points on this line. So when you take the limit, you get 1.

Now consider the line $y=-x$, and take the limit, you get $-1$.

This tells you that the limit cannot exist.

Added : If you want to play around with this idea, try looking at the limits of these functions at $(0,0)$ : $$f(x,y) = \frac{xy}{x^2 + y^2}$$ $$f(x,y) = \frac{xy}{x^3 + y^{3/2}}$$

• so you can choose any y and compare the limit from that direction? Meaning you don't ever have to set a value to x and you can just keep changing one of the two variables (x or y) and compare their value from different directions towards the same point as long as that direction is in the domain of the function? – Raynos Sep 27 '13 at 14:50
• Yes, the idea is that the limit should be the same no matter how you approach $(0,0)$ - be it along a line, or even a curve. The $X$ and $Y$ axis do not play any significant role here. – Prahlad Vaidyanathan Sep 27 '13 at 14:52

You cannot set $x=0$ because the function is not defined there, and the same goes with $y$. Why not look at $x=y$ and $x=-y$then take $x\to 0$?

• ahh I dont understand how everyone is coming up with these substitutions..How would I know which values would be in the domain of f and that they would work? – Raynos Sep 27 '13 at 14:48
• @Raynos Do you know how the function $$f(x)=\dfrac{x}{|x|}$$ behaves? We're now looking at something similar, which is $g(x,y)=f(xy)=\frac{xy}{|xy|}$. It is natural to want to recover a one variable function we know how to deal with by setting $y=x$. This gives a limit of $1$, and setting $x=-y$ gives a limit of $-1$, similar to what happens to $f(x)$ originally, in the $1$-dimensional case. – Pedro Tamaroff Sep 27 '13 at 14:51

No, you can't do that. Try instead $$\lim_{x \to 0} f(x,x)$$ and $$\lim_{x \to 0} f(x,-x).$$