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Math Help - Limit problem

  1. #1
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    Limit problem

    How do you show that the limit as x approaches negative infinity of ex|x|a = 0?

    a > 0
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  2. #2
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    Re: Limit problem

    This thread on math.stackexchange.com may be helpful.
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  3. #3
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    Re: Limit problem

    Well in a previous problem they asked me to show that a) limx->0 xaln(x) = 0 (a > 0)

    That was pretty straight forward using L'hopital's rule on limx->0 ln(x) / x-a

    But I'm having troubles with the followup question b) limx->- inf |x|aex​ = 0

    They say: by putting x = ln (t) in a), show b)

    (?)
    Last edited by Cinnaman; October 24th 2012 at 04:14 AM.
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  4. #4
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    Re: Limit problem

    We have the chain rule for limits, which says that for c,d,e\in\mathbb{R}\cup\{-\infty,\infty\}, if

    1. \lim_{y \to d} f(y) = e,
    2. \lim_{x \to c} g(x) = d, and either
      1. f(d)=e or
      2. g does not take the value d near c


    then \lim_{x \to c} f(g(x)) = e. In fact, c, d and e can also have the form +a or -a for a\in\mathbb{R}; in this case, the rule refers to the corresponding one-sided limit.

    We need to prove \lim_{x\to-\infty}e^x|x|^a = 0 where a > 0. Let b = 1 / a.

    We know that \lim_{y\to+0}f(y) = 0 where f(y)=y^b\ln y. Let g(x)=e^x. Then \lim_{x\to-\infty}g(x)=+0 and e^x\ne0 for all x. By the chain rule, \lim_{x\to-\infty}f(g(x))=0 and f(g(x))=e^{bx}x. Therefore, \lim_{x\to-\infty}e^{bx}|x|=+0.

    Next, let h(z)=z^a. Then \lim_{z\to+0}h(z)=0. By the chain rule again, \lim_{x\to-\infty}h(e^{bx}|x|)=0 and h(e^{bx}|x|)=e^{abx}|x|^a=e^x|x|^a.
    Thanks from Cinnaman
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