# Limits in two variables

1. Feb 26, 2006

### Stevecgz

$$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} xy\ln{xy}$$

I beleive this limit does not exist since the natural log of zero is undefined.

$$\lim_{\substack{x\rightarrow 1\\y\rightarrow 2}} \frac{xy - 2x - y + 2}{x^2 + y^2 - 2x - 4y + 5}$$

I am uncertain how to find this limit. I've tried some different algebraic manipulations but I cannot find a form for which the denominator is not zero.

Any help is appreciated, thanks.

Steve

2. Feb 26, 2006

### d_leet

For the second one, first take the limit as y goes to 2 and then take the limit as x goes to 1, after dealing with y though it should be pretty easy.

3. Feb 26, 2006

### Stevecgz

Ok, I took the limit as y goes to two and the limit as x goes to one and I got a limit of 0 for both of those. I also took the limit for y = x and got zero. I now suspect the limit is zero, but how do I make sure the limit is zero?

4. Feb 26, 2006

### d_leet

Don't take the limits simultaneously, first find the limit as y goes to 2, then look at what you have before you even take the limit as x goes to 1. You could also do it the other way around and of course you get the same limit, but do one limit before the other either do x or y first but not both and then look at what you have before taking the second limit.

5. Feb 26, 2006

### Stevecgz

Ok, when I do that I see that the numerator is going to be zero weather I just take the limit as y goes to 2 or I just take the limit as x goes to 1. I see that this proves that the limit is zero if the limit is approaced along the x-axis or along the y-axis, but there are still many other ways that the limit may be approached, right?

6. Feb 26, 2006

### d_leet

I think so, I'm getting a little bit shaky on what i remember about calc 3, I really need to review some of this stuff, but yeah I think you can approach it along any path in the plane and as long as the limit is the same for any path you take then it exists, however if any two paths you take give you diofferent limits then limit does not exist because it does not tend to a unique value.

7. Feb 26, 2006

### benorin

This may help, notice that

$$\lim_{\substack{x\rightarrow 1\\y\rightarrow 2}} \frac{xy - 2x - y + 2}{x^2 + y^2 - 2x - 4y + 5} = \lim_{\substack{x\rightarrow 1\\y\rightarrow 2}} \frac{(x-1)(y-2)}{(x-1)^2 + (y-2)^2} = \lim_{\substack{u\rightarrow 0\\v\rightarrow 0}} \frac{uv}{u^2 + v^2}$$

where the substitution is permissable since if $$x\rightarrow 1,y\rightarrow 2,$$ then $$u:=x-1\rightarrow 0,v:=y-2\rightarrow 0.$$

Last edited: Feb 26, 2006
8. Feb 26, 2006

### benorin

Now we wish to prove the conjecture that the value of the limit is 0:

and hence we require that

$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that }0<\sqrt{u^2+v^2}<\delta\Rightarrow \left|\frac{uv}{u^2 + v^2}-0\right|<\epsilon$$

9. Feb 26, 2006

### Stevecgz

Thanks guys, I appreciate it. I hadn't thought of the substitution, that makes the proof much easier.

Anyone know about the first one? Am I correct that the limit dne since ln(0) is undefined?

10. Feb 26, 2006

### benorin

That may not be the case, since, for example, along x=y, the first limit is

$$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} xy\ln{xy}=\lim_{x\rightarrow 0} x^2\ln{x^2}=\lim_{x\rightarrow 0} \frac{2\ln{x}}{\frac{1}{x^2}}=\lim_{x\rightarrow 0} \frac{\frac{2}{x}}{-\frac{2}{x^3}}=\lim_{x\rightarrow 0} -x^2=0$$

by l'Hospital's Rule.

11. Feb 26, 2006

### benorin

No, wait, you are likely right, since ln(xy) isn't even defined (well, it's not real) if xy<0.

12. Feb 26, 2006

### AKG

So? 1/0 is undefined, but:

$$\lim _{x \to 0}x\frac{1}{x} = 1$$

They're not asking you to compute the log of 0, they're asking you to compute the limit. Note that the function:

$$(x, y) \mapsto xy\ln(xy)$$

is only defined on $\{(x,y)\ |\ xy > 0\}$. Among other things, this means that approaching along the lines y=0 or x=0 is not permitted. Note that you may as well just look at:

$$\lim _{r \to 0^+}r\ln(r)$$

Can you see why?

Last edited: Feb 26, 2006
13. Feb 26, 2006

### Hurkyl

Staff Emeritus
Textbooks are very unclear on technicalities such as this. I think the intent of most textbooks is that such an expression is supposed to be ill-defined. (Because the argument isn't a function on a punctured disk about the origin)

14. Feb 27, 2006

### Stevecgz

I sopose I'm still a little unclear. In the case of:

$$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} xy\ln{xy}$$

if the limit is approached along the x-axis or along the y-axis it is undefined, so that alone proves that the limit does not exist right?

Am I correct in saying that to prove a limit doesn't exist you only need to find a single path for which is does not exist, or find two paths for which the limit is different, but to prove a limit does exist you must prove it exist (and is the same) on all possible paths?

15. Feb 27, 2006

### Hurkyl

Staff Emeritus
There are two cases:

(1) You're supposed to restrict your attention to only the pairs (x,y) near the origin for which the argument is defined.
(2) You're supposed to proclaim the limit ill-defined, because its argument is undefined near the origin.

And IMHO it's unclear which question is being asked.

16. Feb 27, 2006

### AKG

I think it's reasonable to only look at paths that stay within {(x,y) | xy > 0}, in which case, as I previously mentioned, the x-axis and y-axis are not possible paths.

17. Feb 27, 2006

### VietDao29

Are you sure that limit is 0???
Let's approach it along the u-axis: v = 0, so the limit is 0, but if you approach it along the path u = v, then the limit is 1 / 2. Hence, the limit does not exist.
Thus, that conjecture is wrong...

18. Feb 27, 2006

### benorin

Good point, thanks.

19. Feb 27, 2006

### HallsofIvy

I wrote a whole paragraph disagreeing with you, then suddenly realized what you meant: it isn't just that xy ln(xy) is not defined at (0,0) but that it is not defined at any (x, 0) or (0, y) and so is not defined in a neighborhood of (0, 0). Very good point. I would be inclined to say the limit does not exist.

20. Feb 27, 2006

### Stevecgz

Thank you to all for the replys, it is appreciated.

Steve