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Math Help - Limits

  1. #1
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    Limits

    How can you show that the limit of this as t_1, t_2 approaches 0 is not 0...?

    lim_{t \to 0} \left( \frac{t_1 t_2^4}{t_1^2 + t_2^8} \right)



    I understand why it doesn't approach 0 since t_1^2 + t_2^8 approaches 0 quicker than t_1 t_2^4, but how can it be shown more formally?

    I'd really appreciate the help. Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by WWTL@WHL View Post
    How can you show that the limit of this as t_1, t_2 approaches 0 is not 0...?

    lim_{t \to 0} \left( \frac{t_1 t_2^4}{t_1^2 + t_2^8} \right)



    I understand why it doesn't approach 0 since t_1^2 + t_2^8 approaches 0 quicker than t_1 t_2^4, but how can it be shown more formally?

    I'd really appreciate the help. Thanks.
    I think this is a crude way of saying

    \lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^8}

    Is that right?
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  3. #3
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    Yes, sorry if that wasn't clear. Any ideas?
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by Mathstud28 View Post
    I think this is a crude way of saying

    \lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^8}

    Is that right?
    Yes, but let's simplify this by letting x=y since they both approach the same limit.

    We now have:

    \lim_{x \to 0}\frac{x^5}{x^2+x^8}

    \lim_{x \to 0}\frac{x^3}{1+x^6} = \frac{0}{1} = 0
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by colby2152 View Post
    Yes, but let's simplify this by letting x=y since they both approach the same limit.

    We now have:

    \lim_{x \to 0}\frac{x^5}{x^2+x^8}

    \lim_{x \to 0}\frac{x^3}{1+x^6} = \frac{0}{1} = 0
    Now repeat with y= x^{1/4} and let x go to zero from the right, since for a limit to exist it must give the same result however you approach the point in question.

    CB
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