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Thread: Finding the limit

  1. #1
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    Finding the limit

    Using polar corrdinates, evaluate $\displaystyle \lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^2+y^3}$

    so $\displaystyle x=r\sin\theta$, $\displaystyle y=r\cos\theta$

    $\displaystyle \frac{x^3+y^3}{x^2+y^3}=\frac{r(\sin^3\theta +\cos^3\theta)}{\sin^2\theta +\cos^2\theta}$

    Since $\displaystyle \left|\frac{r(\sin^3\theta +\cos^3\theta)}{\sin^2\theta +\cos^2\theta}\right|\leq r$ hence limit=0

    Can I ask why is that so?

    when x,y goes to 0 respectively, which variable goes to 0? r or $\displaystyle \theta$?
    Last edited by mr fantastic; Oct 17th 2009 at 04:03 PM. Reason: Improved the latex for better formatting.
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    Using polar corrdinates, evaluate $\displaystyle lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^2+\color{red}y^3}$
    Did you notice that it is $\displaystyle y^3$ in the denominator?
    Is that a typo?

    If not, the limit does not exist.
    Look at the path along the x-axis and the along the y-axis.
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  3. #3
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    Do you mean

    $\displaystyle \lim_{\small (x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y^2}?$

    If so, then using the Pythagorean identity $\displaystyle \cos^2\theta+\sin^2\theta=1$, we obtain

    $\displaystyle \frac{x^3+y^3}{x^2+y^2}=\frac{r^3(\cos^3\theta+\si n^3\theta)}{r^2(\cos^2\theta+\sin^2\theta)}=\frac{ r^3(\cos^3\theta+\sin^3\theta)}{r^2\cdot 1}=r(\cos^3\theta+\sin^3\theta).$

    Now notice that $\displaystyle r$ is the variable that goes to $\displaystyle 0$ (as we're approaching the origin), and

    $\displaystyle |\cos^3\theta+\sin^3\theta|\le|\cos^3\theta|+|\sin ^3\theta|\le\cos^2\theta+\sin^2\theta=1.$

    Edit: Added more useful inequality.
    Last edited by Scott H; Oct 17th 2009 at 06:24 AM.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Did you notice that it is $\displaystyle y^3$ in the denominator?
    Is that a typo?

    If not, the limit does not exist.
    Look at the path along the x-axis and the along the y-axis.
    Oh yeah it shd be y^2

    Quote Originally Posted by Scott H View Post
    Do you mean



    $\displaystyle \lim_{\small (x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y^2}?$



    If so, then using the Pythagorean identity $\displaystyle \cos^2\theta+\sin^2\theta=1$, we obtain



    $\displaystyle \frac{x^3+y^3}{x^2+y^2}=\frac{r^3(\cos^3\theta+\si n^3\theta)}{r^2(\cos^2\theta+\sin^2\theta)}=\frac{ r^3(\cos^3\theta+\sin^3\theta)}{r^2\cdot 1}=r(\cos^3\theta+\sin^3\theta).$



    Now notice that $\displaystyle r$ is the variable that goes to $\displaystyle 0$ (as we're approaching the origin), and



    $\displaystyle |\cos^3\theta+\sin^3\theta|\le|\cos^3\theta|+|\sin ^3\theta|\le\cos^2\theta+\sin^2\theta=1.$



    Edit: Added more useful inequality.

    Oh so it's actually the r goes to 0 and not the $\displaystyle \theta$?

    Then by squeeze theorm we conclude it goes to 0 right?
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  5. #5
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    Quote Originally Posted by noob mathematician View Post
    Oh so it's actually the r goes to 0 and not the $\displaystyle \theta$?

    Then by squeeze theorm we conclude it goes to 0 right?
    Correct.

    If $\displaystyle \theta$ goes to $\displaystyle 0$, our point approaches the $\displaystyle x$-axis, but not necessarily the origin.
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