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Math Help - Two tricky two-variable limits

  1. #1
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    Two tricky two-variable limits

    I've been working on these two limits all night and can't seem to make any reasonable progress. Any help would be appreciated. Here's the first one:

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

    According to Wolfram Alpha, this limit does not exist. However, I can't seem to find any x,y path that gives any limit other than 0. I tried all lines (x=0, y=0, y=kx), as well as x=y^2 and y=x^2. Any ideas?

    And the second one:

    \lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}

    Here, I know that

    \lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^2+y^2} = \lim_{(x,y) \to (0,0)} \frac{\frac{1}{x^2+y^2}}{e^{\frac{1}{x^2+y^2}}} = \lim_{t \to \infty} \frac{t}{e^t} = 0

    I can't find a way to use this though.

    Thanks!
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  2. #2
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    \displaystyle \lim_{(x,y) \to (0,0)} \frac{xy^2}{2x-y}

    Try approaching on y=x and y=-x.
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  3. #3
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    Please try this
    <br />
x=sin \; (\frac{1}{2}\alpha)<br />

    <br />
y=sin \; \alpha<br />

    In the second may be polar coordinates be useful.
    Last edited by zzzoak; November 11th 2010 at 04:43 PM.
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Try approaching on y=x and y=-x.
    I tried y=kx for all x (except 2 of course, as y=2x is not in the domain), it's always 0. But thanks anyway.

    Quote Originally Posted by zzzoak View Post
    Please try this
    <br />
x=sin \; (\frac{1}{2}\alpha)<br />

    <br />
y=sin \; \alpha<br />
    Thank you very much, that did it! (I got 4, hope that's correct.)

    Quote Originally Posted by zzzoak View Post
    In the second may be polar coordinates be useful.
    I haven't done a limit in polar coordinates before, but as I understand it, this shold work:

    \displaystyle<br />
\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} = \lim_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4(cos^4 \varphi + sin^4 \varphi)} = \lim_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^2} \cdot \frac{1}{r^2(cos^4 \varphi + sin^4 \varphi)}

    The first fraction approaches 0 (per the formula in my original post). However, it's still 0 \cdot \infty, so that doesn't really help me. Any ideas on how to continue?

    Also, could you point me to a good resource on using polar coordinates with limits? I'm not sure that what I did is really OK -- aren't there any conditions that need to be satisfied before I can say that a limit is the same in polar coordinates? Also, I assumed that I can just ignore the angle \varphi, as I'm approaching zero, is that correct?

    Thanks again, you've been a great help!
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  5. #5
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    You can make substitution
    <br />
t=1/r^2<br />

    and get

    <br />
\displaystyle {lim_{t \to \infty} \; \frac{t^2}{e^t}\; max( \varphi )<br />
}<br />
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