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Thread: Limit of a 2-variable function

  1. #1
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    Limit of a 2-variable function

    Hello everyone!

    I am trying to find a limit for the 2-variavle function $\displaystyle f(x,y)=\frac{(x+y)^2}{x^2+y^2}$.

    So right before changing to polar coordinates, I get something like:
    $\displaystyle L=1+\lim_{(x,y) \rightarrow (+\infty ,+\infty)} \, \frac{2xy}{x^2+y^2}$$\displaystyle =1+\lim_{r \rightarrow +\infty} \, \frac{2r^2 \sin \theta \cos \theta}{r^2}$ and thus $\displaystyle L$ is bounded by 0 and 2: $\displaystyle 0<L<2$.

    (1) Does this mean that the limit exists?
    (2) Can we conclude that the numerator and the denomenator are of the same "order"?
    (because the ratio of the numerator and the denominator is a constant)

    Any help is appreciated!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Easier: the limits along $\displaystyle y=x$ and $\displaystyle y=-x$ are different.

    Regards.

    Fernando Revilla
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  3. #3
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    Okay. I may need to add something. Since, x and y are approching $\displaystyle +\infty$ ten the angle $\displaystyle \theta$ is between 0 and 90 degrees so $\displaystyle 1<L<2$.
    You're right, it's much easier, but my approach was to get a limit that is not zero or $\displaystyle \infty$ to determine wether the numerator and denominator are of same order eg. $\displaystyle x^3+\log x $ and $\displaystyle 4x^3-\log 2x -10$ are of same order...
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    I understand. But even if you consider $\displaystyle 0\leq \theta \leq \pi/2$ we have:

    $\displaystyle f(\rho\cos \theta,\rho \sin \theta)=1+\sin 2\theta$

    so, the limit varies according to $\displaystyle \theta$

    Regards.

    Fernando Revilla
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  5. #5
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    If you're going to use polars, note that $\displaystyle \displaystyle \frac{(x + y)^2}{x^2 + y^2} = \frac{x^2 + y^2 + 2xy}{x^2 + y^2}$.

    Using the change of coordinates $\displaystyle \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2$ we find

    $\displaystyle \displaystyle \frac{x^2 + y^2 + 2xy}{x^2 + y^2} = \frac{r^2 + 2r^2\sin{\theta}\cos{\theta}}{r^2} = 1 + 2\sin{\theta}\cos{\theta}$.


    Notice that this does not depend on $\displaystyle \displaystyle r$, so making $\displaystyle \displaystyle r \to \infty$ won't make any difference.

    Since the limit will change over different paths (depending on the value of $\displaystyle \displaystyle \theta$) we can conclude that the limit does not exist.
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