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Math Help - Limits Help

  1. #1
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    Limits Help

    Limit as x goes to negative infinity

    (3x^2 + 1)^-2

    (Last part is raised to the -2 power not minus 2)
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by NRS11 View Post
    Limit as x goes to negative infinity

    (3x^2 + 1)^-2

    (Last part is raised to the -2 power not minus 2)
    like this?

    \lim_{x\to\infty^-}\left(3x^2+1\right)^{-2}

    or

    \lim_{x\to\infty^-}\frac{1}{\left(3x^2+1\right)^{2}}
    Last edited by bigwave; January 11th 2010 at 04:23 PM.
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  3. #3
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    The top one that you mentioned. It is NOT the fraction one.
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  4. #4
    Super Member bigwave's Avatar
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    Cool

    they are both the same thing

    the 2nd is easier to work with
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  5. #5
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    The key is to know that for n>0 is \frac1{t^n}\to0 as t\to+\infty.

    Now, in your limit put t=-x so the limit equals \lim_{t\to+\infty}\frac1{(3t^2+1)^2}, so for large t is \frac{1}{16t^{4}}<\frac{1}{\left( 3t^{2}+1 \right)^{2}}<\frac{1}{9t^{4}}, and the squeeze theorem finishes the problem.
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  6. #6
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    \frac{1}{(3x^2+1)^2}

    is simply a bell-shaped curve with a maximum of 1 at x=0.

    As x approaches + or -infinity, f(x) rapidly goes to zero.
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