1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]\begin{Bmatrix}x=-1+r\cos\theta\\y=3+r\sin\theta\end{matrix}[/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

    jgens

    User Avatar
    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

    LCKurtz

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook