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Math Help - Limit on two variables using the polar form

  1. #1
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    Limit on two variables using the polar form

    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?


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

    \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}

    \displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2\cos\theta\sin\thet  a}{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]}}=

    =\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

    r is always positive, as we defined it.

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

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

    \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}}}}

    Bye there, thanks for posting.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Ulysses View Post
    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?


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

    \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}

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

    =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\thetasen\theta}{\c  os\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}=\displaystyle  \lim_{r \to{0}+}{\displaystyle\frac{cos\thetasen\theta}{r^  2-\cos\theta\sin\theta}}=-1

    r is always positive, as we defined it.

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

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

    \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}}}}

    Bye there, thanks for posting.
    well already at this point you see that limit doesn't exist

     =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\thetasen\theta \sin\theta}{\cos\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}={\displaystyl  e\frac{cos\thetasen\theta \sin\theta}{1+ \cos\theta\sin\theta-2\cos\theta\sin\theta}} ={\displaystyle\frac{cos\thetasen\theta \sin\theta}{1-\cos\theta\sin\theta}}

    because for different values of the  \theta you have different values of limit...
    Last edited by yeKciM; September 12th 2010 at 09:19 AM.
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  3. #3
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    Thanks. I've lost a sine on the way, I've already corrected it, but I think your answer holds. Anyway, as I got the same in the numerator and in the denominator for the first limit, excepting that in the denominator I got the square of the radius and the expression is negative I thought it tended to be -1 when the radius ->0 whats wrong with that?
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Ulysses View Post
    Thanks. I've lost a sine on the way, I've already corrected it, but I think your answer holds. Anyway, as I got the same in the numerator and in the denominator for the first limit, excepting that in the denominator I got the square of the radius and the expression is negative I thought it tended to be -1 when the radius ->0 whats wrong with that?
    \sin ^2 \theta + \cos ^2 \theta = 1

    that's something you miss there so it's not r^2 like you wrote there , so \displaystyle \lim _{r \to anything} of something without r is just that something


    P.S. I edited a little that #2 post
    Last edited by yeKciM; September 12th 2010 at 09:22 AM.
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  5. #5
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    Quote Originally Posted by Ulysses View Post
    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?


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

    \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}

    \displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2\cos\theta\sin\thet  a}{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]}}=

    =\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}}
    Okay, I am with you to here

    =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{r  ^2-\cos\theta\sin\theta}}=-1
    How did the "r" get back into this fraction? (Added: sin^2(\theta)+ cos^2(\theta)= 1, not r^2!)

    From the line above this you can cancel both the " r^2 and cos(\theta)sin(\theta) in the numerator with the same in the denominator getting
    \lim_{r\to 0^+}\frac{1}{(cos(\theta)- sin(\theta)^2}
    which clearly depends on \theta. for example, it \theta= 0, that is 1. If \theta= \pi/4 it is not defined. That alone is enough to tell you that the limit does not exist.

    r is always positive, as we defined it.

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

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

    \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}}}}

    Bye there, thanks for posting.
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  6. #6
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    You're right hallsofIvy, what happenned with that r^2 is that I thought of the sine and the cosine as beeing x and y :P

    I see my mistake now, your help was clarifying. Thank you both.

    Bye there!
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