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L'Hospital's Rule

  • Thread starter endeavor
  • Start date
176
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1. Homework Statement
Find the limit.
[tex]\lim_{x\rightarrow -\infty} x^2e^x[/tex]

2. Homework 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?
 

Answers and Replies

Curious3141
Homework Helper
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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.
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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it would seem like the x term would overcome e^x, and the limit would go to infinity.
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.
 
176
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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?
 
184
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intuition.
Looking at this one
[tex]\lim_{x\rightarrow \infty} \frac{x^2}{e^x}[/tex]

Which one approaches infinity 'faster'?
 
HallsofIvy
Science Advisor
Homework Helper
41,732
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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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