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Homework Help: Limit with two variables

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    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, I've been reading some old threads like https://www.physicsforums.com/showthread.php?t=114567&page=2" (there a limit likely the one I'm threatening here but not exactly the same when they talk about the definition of the neighbors at the point), and 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.

    [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{(x^2+y^2)\ln(x^2+y^2)}=\displaystyle\lim_{r \to{0}+}{(r^2)\ln(r^2)}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\ln(r^2)}{\displaystyle\frac{1}{r^2=\displaystyle\lim_{r \to{0}+}\displaystyle\frac{2r}{r^2
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Sep 13, 2010 #2

    HallsofIvy

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    I am okay to here but how did you get the next term?

    And, of course, [tex]\frac{2r}{r^4}= -r^{-3}[/tex], which has no limit as r goes to 0, not "[tex]-r^3[/tex]".

    Yes, but your calculation is not valid!

    From [tex]\lim_{r\to 0^+}\frac{ln(r^2)}{\frac{1}{r^2}}=\lim_{r\to 0^+} \frac{2ln(r)}{r^{-2}}[/tex] you can use L'Hopital's rule to get [tex]\lim_{r\to 0^+}\frac{2\frac{1}{r}}{-2r^{-3}}= \lim_{r\to 0^+}-r^2= 0[/tex].
     
    Last edited by a moderator: Apr 25, 2017
  4. Sep 13, 2010 #3
    I've used l'hopital
    [tex]\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\ln(r^2)}{\displaystyle\frac{1}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{2r}{r^2}:\displaystyle\frac{-2r}{r^4}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{2}{r}:\displaystyle\frac{-2}{r^3}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{-r^3}{r}}=\displaystyle\lim_{r \to{0}+}{-r^2}[/tex]

    You're right :P
     
  5. Sep 13, 2010 #4
    What you think about this one?

    [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}[/tex]

    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.

    [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\displaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}[/tex]

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

    [tex]\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\cos^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\theta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\theta}{cos^2\theta}[/tex]
    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?
     
  6. Sep 13, 2010 #5
    No, your intuition is off here. The limit of a product equals the product of the limits only when each individual limit actually exists.

    I don't think the limit exists. Try approaching 0 along x = y^2.
     
  7. Sep 15, 2010 #6
    Thanks.
     
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