Page 1 of 2 12 LastLast
Results 1 to 15 of 26

Math Help - Limit with two variables

  1. #1
    Member
    Joined
    May 2010
    Posts
    241

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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)
    Last edited by yeKciM; September 13th 2010 at 11:05 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    Posts
    241
    What you think about this one?

    \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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    What you think about this one?

    \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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2010
    Posts
    241
    Is it because it oscillates? so, it doesn't matter that its multiplied by something that tends to zero?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2010
    Posts
    241
    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 }
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2010
    Posts
    241
    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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?
    please read my last post ....


    \displaystyle \lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}} doesn't exist
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    May 2010
    Posts
    241
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by Ulysses View Post
    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
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5

    \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.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    May 2010
    Posts
    241
    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?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. limit (two variables)
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 7th 2012, 07:37 AM
  2. limit of several variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 18th 2011, 03:55 PM
  3. Limit of a 2 variables function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 14th 2009, 08:05 AM
  4. Limit with Three Variables
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 8th 2009, 10:52 PM
  5. Limit (two variables)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 14th 2007, 05:56 PM

Search Tags


/mathhelpforum @mathhelpforum