Limits: Solving x^3e^(-x^2) with L'Hospital's Rule

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Discussion Overview

The discussion revolves around evaluating the limit of the expression x^3 * e^(-x^2) as x approaches infinity, specifically using L'Hospital's Rule. Participants explore various approaches and reasoning related to the limit's behavior.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about their application of L'Hospital's Rule and seeks clarification on their approach.
  • Another participant clarifies that the limit being evaluated is indeed x^3 * e^(-x^2) as x approaches infinity.
  • Some participants propose that the limit should be zero, arguing that e^(x^2) grows significantly faster than x^3.
  • A participant questions the method of "moving down" x^3 and suggests rewriting e^(-x^2) for clarity.
  • One participant notes that applying L'Hospital's Rule may require multiple iterations, specifically three times, to reach a conclusion.
  • A later reply indicates that after applying L'Hospital's Rule three times, they arrived at a limit expression but challenges the correctness of the denominator derived in the process.

Areas of Agreement / Disagreement

Participants express differing views on the application of L'Hospital's Rule and the resulting limit. There is no consensus on the correct approach or final answer, as some believe the limit is zero while others question the calculations involved.

Contextual Notes

Participants highlight potential misunderstandings in the application of derivatives and the product rule, indicating that the discussion may involve unresolved mathematical steps and assumptions about growth rates.

grothem
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Lim x[tex]\rightarrow[/tex][tex]\infty[/tex] x^3 [tex]\times[/tex] e^(-x^2)

I moved x^3 down to make it a fraction and then I used L'Hospital's rule, but I can't come up with the right answer. Not sure what I'm doing wrong
 
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that looks kind of confusing...

its the limit of x^3 * e^(-x^2) as x approaches infinity
 
well, i guess this is going to be zero, since [tex]e^{x^2}[/tex] grows much faster than x^3
 
What do you mean you "moved it down"?

Regardless, I suggest rewriting [itex]e^{-x^{2}}[/itex] using the facts you know about negative exponents. Probably easier that way.
 
and if you want to use l'hospital's rule, you need to do it three times from what i can see!
 
ah ok...I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0
 
grothem said:
ah ok...I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0

Well the bottom will defenitely not be what u got!
[e^(x^2)]'=(x^2)'e^(x^2)]=[2xe^(x^2)] not take again the derivative and apply the product rule

(uv)'=u'v+v'u
 

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