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I Does this go to 0 for large enough x?

  1. Aug 3, 2016 #1

    ChrisVer

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    I have one question... In general I always thought that the exponential function was "dying" out faster than any other polynomial function, such that:
    [itex]e^{-x} x^a \rightarrow 0[/itex] for [itex]x \rightarrow \infty[/itex].
    [eg this is was used quiet commonly and so I got it as a rule-of-thumb, when deriving wavefunctions for a simple example for the Hydrogen atom]

    However recently I read in a paper that this is not true, and as an illustration of how can that be, they logarithm-ized the function like:
    [itex]\ln (e^{-x} x^n) = -x + n \ln x[/itex] which goes to infinity for [itex] x,n\rightarrow \infty[/itex].
    This I read in here:
    http://arxiv.org/pdf/1108.4270v5.pdf
    in Sec4 (the new paragraph after Eq4.1)

    This has confused me, can someone shred some light?
     
  2. jcsd
  3. Aug 3, 2016 #2

    micromass

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    ##n## is variable. In your example, ##a## is fixed.
     
  4. Aug 3, 2016 #3

    ChrisVer

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    well even if [itex]n[/itex] was not a variable, then the quantity I wrote [itex]-x + n \ln x[/itex] (fixed n), is not really going to zero for large x...

    oops sorry... you want the logarithm to go to infinity.
     
  5. Aug 3, 2016 #4

    Stephen Tashi

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    which is on page 10 of the PDF.

    Your rule-of-thumb is correct about limits involving single variables.

    However, there is a distinction between limits taken with respect to one variable and limits taken with respect to two variables.

    ##\lim_{s\rightarrow \infty} g(s) ## is defined differently than ##\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)##.

    (The wording in the paper is "when both ##n## and ##s## go to ##\infty##".)

    There is a further distinction between the definition of a "double limit" ##\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)## and the definition of the two "iterated limits":
    ##\lim_{s\rightarrow\infty}( \lim_{n\rightarrow\infty} g(s,n))##
    and
    ##\lim_{n\rightarrow\infty}( \lim_{s\rightarrow\infty} g(s,n))##.
     
  6. Aug 4, 2016 #5

    vanhees71

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    The first limit can be evaluated by repeated application of de L'Hospital's rule,
    $$\lim_{x \rightarrow \infty} \frac{x^{a}}{\exp x}=\lim_{x \rightarrow \infty} \frac{a x^{a-1}}{\exp x}=\ldots = \lim_{x \rightarrow \infty} \frac{a(a-1) \ldots (a-n+1) x^{a-n}}{\exp x}=0,$$
    where I made ##n>a-1##.
     
  7. Aug 4, 2016 #6

    Stephen Tashi

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    L'hospitals rule doesn't apply once the numerator becomes constant, so we should stop when that happens.
     
  8. Aug 5, 2016 #7

    vanhees71

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    If ##a \in \mathbb{N}## then for ##n=a## the numerator becomes constant, and then the limit is also shown to be 0. So there's nothing wrong arguing with de L'Hospital in all cases.
     
  9. Aug 5, 2016 #8

    Stephen Tashi

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    I'm just saying L'Hospital's rule does not apply to a case like ##lim_{x\rightarrow\infty} \frac{6}{f(x)}## since L'Hospital's rule only applies when both the numerator and denominator both have a limit of zero or both have infinite limits. Once you get a constant like 6 in the numerator, you have to use a different justification for finding the limit.

    So we shouldn't write a chain of equalities that implies L'Hospital's rule is being applied to the case where the numerator is constant because it is not, in general, true that ##lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = lim_{x\rightarrow\infty} \frac{f'(x)}{g'(x)} =lim_{x\rightarrow\infty} \frac{f"(x)}{g"(x)} = .... = lim_{x\rightarrow\infty} \frac{\frac{d^n}{dx} f(x)}{\frac{d^n}{dx}g(x)} ##, running through an arbitary number ##n## of differentiations.
     
  10. Aug 5, 2016 #9

    vanhees71

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    Of course it works only for limits of the type "##0/0##" or ##\infty/\infty##. Perhaps I was not strict enough in my formulation. I did not mean to use it arbitrarily many times of course but only as long as it is applicable.
     
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