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Math Help - Limits of Curves

  1. #1
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    Limits of Curves

    Find all values of 'alpha' for which the limit exists. For these values, evaluate the limit.

    See the limit below...

    Thanks
    Attached Thumbnails Attached Thumbnails Limits of Curves-math-question.png  
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  2. #2
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    Quote Originally Posted by JoAdams5000 View Post
    Find all values of 'alpha' for which the limit exists. For these values, evaluate the limit.

    See the limit below...

    Thanks
    use the identity 2 \cos^2 \left(\frac{\pi}{4} - \frac{|x|^{\alpha} + |y|^{\alpha}}{2(x^2+y^2)} \right)=1 + \sin \left(\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \right). now consider three cases: \alpha < 2, \ \alpha = 2, \ \alpha > 2. show that if \alpha < 2, the limit doesn't exist. if \alpha=2, then the limit is

    obviously 1+\sin 1. finally if \alpha > 2, then \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to 0 and thus in this case the limit is equal to e.
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  3. #3
    Senior Member Sampras's Avatar
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    Wouldn't <br />
\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty<br />
if  \alpha > 2 ?
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  4. #4
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    Quote Originally Posted by Sampras View Post
    Wouldn't  \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} \to \infty if  \alpha > 2 ?
    no! put x=r\cos \theta, \ y=r \sin \theta. then, since (x,y) \to (0,0), we have r=\sqrt{x^2+y^2} \to 0. thus: 0 \leq \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} =r^{\alpha - 2}(|\cos \theta|^{\alpha} + |\sin \theta|^{\alpha}) \leq 2 r^{\alpha -2} \to 0.
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    Senior Member Sampras's Avatar
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    if x and y were fixed...it would be right?
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    Quote Originally Posted by Sampras View Post
    if x and y were fixed...it would be right?
    you mean if x,y were fixed and \alpha \to \infty, then \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty ? is that what you're asking?
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  7. #7
    Senior Member Sampras's Avatar
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    yes
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  8. #8
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    Quote Originally Posted by Sampras View Post
    yes
    the limit will depend on x,y: if 0 < |x|, |y| < 1, the limit is 0. if |x| = |y| = 1, the limit is 1. if |x| > 1 or |y| > 1, then the limit is infinity. if x = y = 0, the limit obviously doesn't exist.

    to avoid any possible confusion: Sampras' question is kind of off-topic and not related to JoAdams5000's question.
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  9. #9
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    use the identity 2 \cos^2 \left(\frac{\pi}{4} - \frac{|x|^{\alpha} + |y|^{\alpha}}{2(x^2+y^2)} \right)=1 + \sin \left(\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \right). now consider three cases: \alpha < 2, \ \alpha = 2, \ \alpha > 2. show that if \alpha < 2, the limit doesn't exist. if \alpha=2, then the limit is

    obviously 1+\sin 1. finally if \alpha > 2, then \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to 0 and thus in this case the limit is equal to e.
    How did you come up with the identity:   2 \cos^{2} \left(\frac{\pi}{4}- \frac{\theta}{2} \right) = 1 + \sin \theta ? And how did you get  e for  \alpha < 2 ?
    Last edited by Sampras; June 26th 2009 at 09:09 PM.
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