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L'Hospital's Rule

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the limit.
    [tex]\lim_{x\rightarrow -\infty} x^2e^x[/tex]

    2. Relevant equations
    L'Hospital's Rule.

    3. The attempt at a solution
    I rewrite the limit:
    [tex]\lim_{x\rightarrow -\infty} \frac{e^x}{\frac{1}{x^2}}[/tex]
    applying L'Hospital's Rule:
    [tex]\lim_{x\rightarrow -\infty} \frac{e^x}{-2(\frac{1}{x^3})}[/tex]
    But the numerator and denominator still go to zero...If I keep applying L'Hospital's Rule, I still won't get anywhere. It looks like x's exponent will continue to grow; then if I put the x term back in the numerator, it would seem like the x term would overcome e^x, and the limit would go to infinity. However, I graphed the function, and it seems to go to zero. What am I doing wrong?
     
  2. jcsd
  3. Jan 20, 2007 #2

    Curious3141

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    Hint : try rewriting the limit as [tex]\lim_{x\rightarrow \infty} {(-x)}^2e^{-x}[/tex] = [tex]\lim_{x\rightarrow \infty} \frac{x^2}{e^x}[/tex] and apply L'Hopital's rule twice.
     
  4. Jan 20, 2007 #3

    Hurkyl

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    Let this be your first lesson that exponential factors dominate polynomial factors. :smile:

    Anyways, as Curious suggests, try moving the exponential to the bottom instead of the monomial.
     
  5. Jan 20, 2007 #4
    Yeah, that works. It must've slipped my mind that [tex]\lim_{x\rightarrow -\infty} f(x) = \lim_{x\rightarrow \infty} f(-x)[/tex].

    Is there any other way to do it though?
     
  6. Jan 21, 2007 #5
    intuition.
    Looking at this one
    [tex]\lim_{x\rightarrow \infty} \frac{x^2}{e^x}[/tex]

    Which one approaches infinity 'faster'?
     
  7. Jan 21, 2007 #6

    HallsofIvy

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    For any positive n, a> 1, ax goes to infinity faster than xn so the fraction you have goes to 0. We say that ax "dominates" xn as Hurkyl said.


    For example, for large enough x, 1.0000001x is larger than x100000000.
     
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