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Math Help - Limit of Vector Valued Function

  1. #1
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    Limit of Vector Valued Function

    How would I approach a question such as:
    find the limits:
    1-lim{ (x^4 + y^4) / (x^2+ y^2)} as (x,y) tends to (0,0)
    2-lim{ xy^2 / (x^2 + y^4)} as (x,y) tends to (0,0)

    Thanks
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  2. #2
    Junior Member rubix's Avatar
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    edit: nvm what i just said.

    polar coordinate that Twig showed is the way to go. It drops you down from two variables (x, y) to one (r).
    Last edited by rubix; April 17th 2009 at 05:14 AM.
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  3. #3
    Senior Member Twig's Avatar
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    Hi

    1st one:

     \lim_{(x,y) \rightarrow (0,0)} \frac{x^4 + y^4}{x^2+y^2}

    Use polar coordinates,  x = r cos(\theta) \, \,  y = r sin(\theta)

    Gives us:  \frac{r^4(cos^{4}(\theta) + sin^{4}(\theta))}{r^2} = r^2(cos^{4}(\theta) + sin^{4}(\theta))

    And here  |(cos^{4}(\theta) + sin^{4}(\theta))| \leq 2 \, \Rightarrow r^2(cos^{4}(\theta) + sin^{4}(\theta)) \mbox{ goes to zero as r approaches zero}

    2nd one:

    If we let  x = 0 , then the expression is already zero, which approaches zero.

    But if we let  x = y^2 \, \Rightarrow \frac{y^4}{2y^4} \, \mbox{ which approaches } \frac{1}{2}

    What does this tell you?
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  4. #4
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    Quote Originally Posted by Twig View Post
    Hi

    1st one:

     \lim_{(x,y) \rightarrow (0,0)} \frac{x^4 + y^4}{x^2+y^2}

    Use polar coordinates,  x = r cos(\theta) \, \,  y = r sin(\theta)

    Gives us:  \frac{r^4(cos^{4}(\theta) + sin^{4}(\theta))}{r^2} = r^2(cos^{4}(\theta) + sin^{4}(\theta))

    And here  |(cos^{4}(\theta) + sin^{4}(\theta))| \leq 2 \, \Rightarrow r^2(cos^{4}(\theta) + sin^{4}(\theta)) \mbox{ goes to zero as r approaches zero}

    2nd one:

    If we let  x = 0 , then the expression is already zero, which approaches zero.

    But if we let  x = y^2 \, \Rightarrow \frac{y^4}{2y^4} \, \mbox{ which approaches } \frac{1}{2}

    What does this tell you?
    Hi
    It tells me there is no limit at the origin since there are points close to the origin which the function takes the value 1/2 and not 0 ?

    Why use polar coordinates on the first one?
    Also, are there any pointers you can advise on how to approach these questions, i.e. finding the value of a limit at a point(for vector valued fns)?
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  5. #5
    Senior Member Twig's Avatar
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    hi

    I used polar coordinates because I find it convenient and relatively easy here to show that no matter what angle we approach by, the function tends to zero.
    Im not that familiar with vector valued functions to be honest, I have only done these types of limits when  f: \mathbb{R}^{2} \rightarrow \mathbb{R} .

    But I would imagnine that both limits have to exist for a vector to approach some point so to speak.
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