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Thread: Limits Help

  1. #1
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    Limits Help

    Limit as x goes to negative infinity

    $\displaystyle (3x^2 + 1)^-2$

    (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

    $\displaystyle (3x^2 + 1)^-2$

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

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

    or

    $\displaystyle \lim_{x\to\infty^-}\frac{1}{\left(3x^2+1\right)^{2}}$
    Last edited by bigwave; Jan 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 $\displaystyle n>0$ is $\displaystyle \frac1{t^n}\to0$ as $\displaystyle t\to+\infty.$

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

    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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