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Math Help - Multi-variable limit

  1. #1
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    Multi-variable limit

    I'll appreciate it if someone can throw me in the right direction on this one:
    prove:

    Multi-variable limit-untitled.png

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SpinRat View Post
    I'll appreciate it if someone can throw me in the right direction on this one:
    prove:

    Click image for larger version. 

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    Thanks
    For the limit to exist it must be the same no matter which direction we approach it. I'd recommend approaching the origin along the lines y = x and y = -x.

    y = x:
    \displaystyle \lim_{(x,y) \to (0,0)} \frac{sin(x^2)~sin^3(y)}{x^6 + y^4}

    \displaystyle = \lim_{x \to 0} \frac{sin(x^2)~sin^3(x)}{x^6 + x^4}

    This is in the form to use l'Hopital's rule, but its a mess to do it this way. Expanding a bit:
    \displaystyle = \lim_{x \to 0} \left ( \frac{sin(x^2)}{x^2} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x^2 + 1} \right )

    \displaystyle = \lim_{x \to 0}  \frac{sin(x^2)}{x^2} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}

    \displaystyle  = 1 \cdot 1 \cdot 1 \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}

    This last limit is 0. y = -x is done in the same way.

    -Dan
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  3. #3
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    But that doesn't prove the limit.
    Actually, in the first section of this question we were asked to show that for a different function the limit exist when approaching through every line {(x,y) | y=ax or x=0}, yet the limit doesnt exist.

    this was the other one:
    Multi-variable limit-untitled.png
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  4. #4
    Super Member girdav's Avatar
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    Use |\sin t|\leq |t| and ab\leq \frac {a^2+b^2}2.
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  5. #5
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    been at it all day, still no go. I think I'm missing something obvious. plz help...
    Multi-variable limit-0.png
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