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Math Help - 2-variable limit

  1. #1
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    2-variable limit

    Can someone please help me to show that:

    [x^5 + y^2]/[x^4 + |y|] -> 0 as (x,y) -> (0,0)
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  2. #2
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    Re: 2-variable limit

    Quote Originally Posted by marqushogas View Post
    Can someone please help me to show that:

    [x^5 + y^2]/[x^4 + |y|] -> 0 as (x,y) -> (0,0)
    Maybe converting to polars might help?

    \displaystyle x = r\cos{\theta} and \displaystyle y = r\sin{\theta}, so

    \displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)}\frac{x^5 + y^2}{x^4 + |y|} &= \lim_{r \to 0}\frac{r^5\cos^5{\theta} + r^2\sin^2{\theta}}{r^4\cos^4{\theta} + |r||\sin{\theta}|} \\ &= \lim_{r \to 0}\frac{r^2\left(r^3\cos^5{\theta} + \sin^2{\theta}\right)}{|r|\left(|r|r^2 \cos^4{\theta} + | \sin{\theta} |  \right)} \\ &= \lim_{r \to 0}\frac{|r|\left(r^3\cos^5{\theta} + \sin^2{\theta} \right)}{|r|r^2\cos^5{\theta} + | \sin{\theta} |} \\ &= \frac{0\left( 0 + \sin^2{\theta} \right)}{0 + | \sin{\theta} |} \\ &= \frac{0}{|\sin{\theta}|} \\ &= 0 \end{align*}
    Last edited by Prove It; September 6th 2011 at 10:43 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: 2-variable limit

    Quote Originally Posted by Prove It View Post
    = \frac{0}{|\sin{\theta}|}= 0
    \frac{0}{|\sin{\theta}|}= 0 \Leftrightarrow \sin \theta\neq 0 and \sin \theta =0 iff (x,y) belongs to the set S=\{(x,0):x\neq 0\} . Now, f(x,y)=x on S and the limit of f(x,y) along S for (x,y)\to (0,0) is also 0 .
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