Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of one exponential function

  1. May 31, 2010 #1
    Hello,
    How can I find the limit of [tex]x^{k}\mathrm{e}^{-ax^n}[/tex] for [itex]x\to +\infty[/itex].
    a,k,n are positive reals.
    I even have troubles solving it for k=1, a=1, n=2.
    Any hint?
     
  2. jcsd
  3. May 31, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    e^x grows faster than any power of x. That's the really important thing here.
     
  4. May 31, 2010 #3
    Thanks, I've heard that explanation somewhere but:

    - Can you please formally specify what you mean by "grows faster".

    - And also, is there a theorem that proves something about the fact that if one function grows faster than the other, it somehow "wins".
    Here x goes to infinity, and exp goes to zero. It's not trivial to deduce that the limit is zero.
     
  5. May 31, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Now that's a good question... I don't know the specific theorem (although I know there is one), but I mean to say that, for example, and this is actually not exactly formal anyway, that e^-x will fall to zero much faster than x^n will climb to infinity. I'm not sure how to put that formally...
     
  6. May 31, 2010 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    The formal statement would be something like:

    "Let two functions f(x) and g(x) be given, and let a be some common boundary point of their domain. We say that f(x) tends to zero faster than g(x) if the limit of f(x) / g(x) as x tends to a is zero."

    If you want to get really formal, you may want to check out the "Big oh" (and related little o) notation.
     
  7. May 31, 2010 #6

    Cyosis

    User Avatar
    Homework Helper

  8. May 31, 2010 #7
    Now I am thinking, could it be this way?
    We consider [itex]f(x)=x^k[/itex] and [itex]g(x)=\mathrm{e}^{ax^n}[/itex].

    Consider [itex]f(x)/g(x)[/itex] and apply the De L'Hopital's rule?
     
  9. May 31, 2010 #8

    Cyosis

    User Avatar
    Homework Helper

    Yes that will work.
     
  10. Jun 1, 2010 #9

    Landau

    User Avatar
    Science Advisor

    It's very easy to prove
    [tex]\lim_{x\to\infty}e^{-x}x^n=0.[/tex]
    Just consider the power series expansion of exp.
    [tex]\frac{e^x}{x^n}=\frac{1+x+\frac{x^2}{2!}+...}{x^n}=\underbrace{\frac{1}{x^n}}_{\to 0}+\underbrace{\frac{1}{x^{n-1}}}_{\to 0}+\underbrace{...}_{\to 0}+\frac{1}{n!}+\frac{x}{(n+1)!}+...\ \ \to\infty.[/tex]
     
    Last edited: Jun 1, 2010
  11. Jun 1, 2010 #10

    Char. Limit

    User Avatar
    Gold Member

    I thought the Taylor series would unlock it... thanks for the help.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook