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Homework Help: Limits using polar forms

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, I've made a double limit using the polar forms. The thing is the limit is wrong, I've made a plot, and then I saw that the limit doesn't exist, and what I wanna know is what I'm reasoning wrong, and some tips to get a deeper comprehension on this limits, and on what I am doing. For the last one I wanna know the limit value, I think it doesn't exists neither. Is it because the sine and cosine oscillates?

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

    [tex]\begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}[/tex]

    [tex]\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2\cos\theta\sin\theta}{r^2\cos\theta\sin\theta+(r\cos\theta-r\sin\theta)^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2cos\theta\sin\theta}{r^2[\cos\theta\sin\theta(cos\theta-\sin\theta)^2]}}=[/tex]

    [tex]=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{\cos\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{r^2-\cos\theta\sin\theta}}=-1[/tex]

    r is always positive, as we defined it.

    [tex]\displaystyle\lim_{(x,y) \to{(-1,3)}}{\displaystyle\frac{\sqrt[ ]{x+y-2}}{(x+1)^2+(y-3)^2}}[/tex]


    [tex]\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{r}\sqrt[ ]{\cos\theta+\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{\cos\theta+\sin\theta}}{r^{\frac{3}{2}}}}[/tex]

    Bye there, thanks for posting.
    Last edited: Sep 12, 2010
  2. jcsd
  3. Sep 12, 2010 #2


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    Gold Member

    I'm not going to comment on your work before this (mostly because I'm not really qualified to do so), but sin2(x)+cos2(x)=1, not r2.
  4. Sep 12, 2010 #3
    Right, thanks. I thought of it as beeing x and y :P
  5. Sep 12, 2010 #4


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    Science Advisor
    Homework Helper
    Gold Member

    I didn't check your work but I have a question for you. If a limit in fact doesn't exist, it is usually easier to find a couple of paths along which you get different values. Have you tried that or are you using polar coordinates because it is required for the problem?

    In either case I would suggest looking at the lines y = x or y = -x in either polar or rectangular form.
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