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Math Help - Limit of a 2 variables function

  1. #1
    MHF Contributor arbolis's Avatar
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    Limit of a 2 variables function

    Say whether or not the limit of f(x,y) when (x,y) tends to (0,0) exists.
    f(x,y)=\frac{xy}{x^2-y^2}-2y.
    My attempt : \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0.

    \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty. As both limits are different, the limit doesn't exist.
    However I'm unsure I didn't make an error by taking y=x^2 for the second limit. Since when x=1, this doesn't even make sense to take such an y.
    The domain of f is any (x,y) \in \mathbb{R}^2 such that x\neq \pm y.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    Say whether or not the limit of f(x,y) when (x,y) tends to (0,0) exists.
    f(x,y)=\frac{xy}{x^2-y^2}-2y.
    My attempt : \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0.

    \color{red}\lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty.
    i don't think that's true. i get that the limit is also 0 along y = x^2

    As both limits are different, the limit doesn't exist.
    However I'm unsure I didn't make an error by taking y=x^2 for the second limit. Since when x=1, this doesn't even make sense to take such an y.
    The domain of f is any (x,y) \in \mathbb{R}^2 such that x\neq \pm y.
    your method is sound though. note that if we approach (0,0) along x = 2y we get -2/3 as the limit. this means the limit doesn't exist

    (the curve you approach the origin must actually go through the origin. the line x = 1 does not.)
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  3. #3
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    Quote Originally Posted by arbolis View Post
    Say whether or not the limit of f(x,y) when (x,y) tends to (0,0) exists.
    f(x,y)=\frac{xy}{x^2-y^2}-2y.
    My attempt : \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0.

    \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty. As both limits are different, the limit doesn't exist.
    However I'm unsure I didn't make an error by taking y=x^2 for the second limit. Since when x=1, this doesn't even make sense to take such an y.
    The domain of f is any (x,y) \in \mathbb{R}^2 such that x\neq \pm y.
    If the given function approaches (0, 0) along the line y = mx ( m \neq \pm 1), the limiting value is \frac{m}{1 - m^2}. So clearly the limiting value depends on how you approach (0, 0) ....
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Jhevon
    i don't think that's true. i get that the limit is also 0 along y = x^2
    I get it now, thanks.
    Quote Originally Posted by mr fantastic
    If the given function approaches (0, 0) along the line y = mx (), the limiting value is . So clearly the limiting value depends on how you approach (0, 0) ....
    I wasn't aware of this! I mean I didn't realize it. This holds for this particular example, right? If it hold for all example then no 2 variable functions would have limit in (0,0), right? I need to clear this doubt!
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    Quote Originally Posted by arbolis View Post
    This holds for this particular example, right? If it hold for all example then no 2 variable functions would have limit in (0,0), right? I need to clear this doubt!
    For this particular problem. Not all.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Plato View Post
    For this particular problem. Not all.
    Thanks, makes sense. I was a bit scared for a moment.
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  7. #7
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    There are, after all, some limits that really exist at (0,0).
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  8. #8
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    There are, after all, some limits that really exist at (0,0).
    Ahah yes.
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