# Thread: Limit with two variables

1. ## Limit with two variables

Hi, I got this other limit here, which I think I've solved rightly. I've used polar coordinates, and I've found a limit that doesn't depends on theta, but I've got some doubts. Since I've looked at many trajectories and everything indicates the solution its fine, I'm not sure about if I'm missing something.

So here it is.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{(x^2+y^2)\ln(x^2+y^2)}=\displaystyle\l im_{r \to{0}+}{(r^2)\ln(r^2)}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\ln(r^2)}{\displaystyl e\frac{1}{r^2}}}=\displaystyle\lim_{r \to{0}+}\displaystyle\frac{2r}{r^2}:\displaystyle\ frac{-2r}{r^4}}=\displaystyle\lim_{r \to{0}+}{-r^3}=0$

So, is this limit right for any trajectory?

2. Originally Posted by Ulysses
Hi, I got this other limit here, which I think I've solved rightly. I've used polar coordinates, and I've found a limit that doesn't depends on theta, but I've got some doubts. Since I've looked at many trajectories and everything indicates the solution its fine, I'm not sure about if I'm missing something.

So here it is.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{(x^2+y^2)\ln(x^2+y^2)}=\displaystyle\l im_{r \to{0}+}{(r^2)\ln(r^2)}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\ln(r^2)}{\displaystyl e\frac{1}{r^2}}}=\displaystyle\lim_{r \to{0}+}\displaystyle\frac{2r}{r^2}:\displaystyle\ frac{-2r}{r^4}}=\displaystyle\lim_{r \to{0}+}{-r^3}=0$

So, is this limit right for any trajectory?
$\displaystyle \lim_{r \to 0 } (r^2 ) \ln {r^2}$

$t = \frac {1}{r}$

$\displaystyle \lim_{t \to \infty} \frac {\ln {\frac {1}{t^2} }} {t^2 } = LR = \lim _{t\to \infty} - \frac {1}{t^2} = 0$

so it's true

P.S. LR == L'Hospital's rule
Edit: sorry (had have problems with computer so needed to edit twice)

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

I think the limit is zero, but I'm having some trouble with its demonstration. I think I've already demonstrated it, but I'm not completely sure.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\d isplaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}$

I've tried to solve the part on the right using polar coordinates, and I get to

$\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim _{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\c os^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\ cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\t heta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\t heta}{cos^2\theta}$
Which means this limit doesn't exists, but as it is multiplied by something that tends to zero, the limit should be zero. Is this right?

4. Originally Posted by Ulysses

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

I think the limit is zero, but I'm having some trouble with its demonstration. I think I've already demonstrated it, but I'm not completely sure.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\d isplaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}$

I've tried to solve the part on the right using polar coordinates, and I get to

$\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim _{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\c os^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\ cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\t heta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\t heta}{cos^2\theta}$
Which means this limit doesn't exists, but as it is multiplied by something that tends to zero, the limit should be zero. Is this right?
this one doesn't exist

5. Is it because it oscillates? so, it doesn't matter that its multiplied by something that tends to zero?

6. Originally Posted by Ulysses
Is it because it oscillates? so, it doesn't matter that its multiplied by something that tends to zero?
start from the begin with polar coordinates .... and you'll end up with something without "r" but limit is r->0 so it doesn't do nothing to the expresion which are functions $\cos \theta , \sin \theta , ...$ which for different values of the $\theta$ have different result ....

that's why limit don't exist

if you have anything let's say

$\displaystyle \lim_{r\to 0 } \frac {\cos \theta \sin \theta }{ \sin \theta + 2 } = \frac {\cos \theta \sin \theta }{ \sin \theta + 2 }$

it's just example ... not that one up there

whenever you get the function as result of limit ... than that limit don't exist

7. Okay, but lets say that the expression is multiplied by g(x), which tends to zero, it doesn't matter?

e.g.
$\displaystyle \lim_{x\to 0 }g(x) \displaystyle \lim_{r\to 0 } \frac {\cos \theta \sin \theta }{ \sin \theta + 2 } = \displaystyle \lim_{x\to 0 }g(x)\frac {\cos \theta \sin \theta }{ \sin \theta + 2 }$

8. Originally Posted by Ulysses
Okay, but lets say that the expression is multiplied by g(x), which tends to zero, it doesn't matter?

e.g.
$\displaystyle \lim_{x\to 0 }g(x) \displaystyle \lim_{r\to 0 } \frac {\cos \theta \sin \theta }{ \sin \theta + 2 } = \displaystyle \lim_{x\to 0 }g(x)\frac {\cos \theta \sin \theta }{ \sin \theta + 2 }$
wait ....

if you have

$\displaystyle \lim_{r\to 0 } r\cdot \frac {\cos \theta \sin \theta }{ \sin \theta + 2 } =0$

than it exists

you wrote there something that reminds me to simultaneous and successive limits and that is something else ....

when you have successive limits

$\displaystyle L_{12} = \lim _{x \to 0 } \lim_{y \to 0} f(x,y)$

$\displaystyle L_{21} = \lim _{y \to 0 } \lim_{x \to 0} f(x,y)$

and if they are equal ....

$\displaystyle L_{12} = L_{21}$

than you have chance that function $f(x,y)$ have limit .... but only if simultaneous limit exists

$\displaystyle L= \lim _{x\to 0 , y \to 0} f(x,y)$

9. Ok. Lets recapitulate. In the example above I got this:

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\d isplaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}$

So, 6x tends to zero, but I don't know what happends with the limit at the right. So what I've made use of the polar coordinates to see what happens with it.

So I got:

$\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim _{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\c os^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\ cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\t heta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\t heta}{cos^2\theta}$

And then: $\displaystyle\frac{\sin^2\theta}{cos^2\theta}\disp laystyle\lim_{(x,y) \to{(0,0)}}{6x}=0$ Right?

10. Originally Posted by Ulysses
Ok. Lets recapitulate. In the example above I got this:

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\d isplaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}$

So, 6x tends to zero, but I don't know what happends with the limit at the right. So what I've made use of the polar coordinates to see what happens with it.

So I got:

$\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim _{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\c os^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\ cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\t heta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\t heta}{cos^2\theta}$

And then: $\displaystyle\frac{\sin^2\theta}{cos^2\theta}\disp laystyle\lim_{(x,y) \to{(0,0)}}{6x}=0$ Right?

$\displaystyle \lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$ doesn't exist

11. All right, I'll do so.

But I've just used the properties of limits, the one that says that the limit of the product between two functions its equal to the product of the limits (when $R^2->R$). Thats what I was trying to do.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

I think I've made a mess while thinking of all, Cartesian and polar forms together.

Thanks.

12. Originally Posted by Ulysses
Damn it, this is confusing.
no it's not ... do your problem from the start with polar .... don't take out anything ... and you'll see that doesn't exist

13. Originally Posted by Ulysses
All right, I'll do so.

But I've just used the properties of limits, the one that says that the limit of the product between two functions its equal to the product of the limits (when $R^2->R$). Thats what I was trying to do.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

I think I've made a mess while thinking of all, Cartesian and polar forms together.

Thanks.
look at it this way (don't know do if you are familiar with it but pay attention)

$\displaystyle L_{12} = \lim _{x \to 0 } \lim_{y \to 0} \frac{6xy^2}{x^2+y^4} = \lim _{x \to 0 } \frac{0}{x^2+0} = 0$

$\displaystyle L_{21} = \lim _{y \to 0 } \lim_{x \to 0} \frac{6xy^2}{x^2+y^4} = \lim _{x \to 0 } \frac{0}{0+y^4} = 0$

so there is chance that this exists .... because successive limits exist and they are equal .... (if they were not than you equal, or one exists and another don't ... or something like that ... that combinations.... than you know limit don't exist )

now you go with simultaneous limit to check if exists ...

$\displaystyle L= \lim _{x\to 0 , y \to 0} \displaystyle \frac {6xy^2}{x^2+y^4} = \lim _{r \to 0 } \frac {6 r^3 \cos \theta \sin^2 \theta}{r^2(\cos^2 \theta + r^2 \sin^4 \theta)} =$

$\displaystyle = \lim _{r \to 0 } \frac {6 r \cos\theta \sin^2\theta} {\cos^2 \theta +r^2\sin^4\theta}= ...$

Edit : sorry that it wasn't to the end there ... i wrote it and have errors to preview and corrected them 3 or 4 times and didn't get to display equation... so i delete to point that will show image... hope you can do the rest by yourself

14. $\displaystyle \lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

why all this silly polar stuff, let $x=my^2$ and use two different m's.

15. Do you use any criteria to realize which "direction" to take? I mean, I've tryied with many direction till I get with the parabola, but I don't know, maybe using sections or something you can "see" which direction to take. Is there any criteria for that, or just intuition when watching at the equation?

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