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Limit problem

  1. Feb 24, 2008 #1
    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
     
  2. jcsd
  3. Feb 24, 2008 #2
    that looks kind of confusing...

    its the limit of x^3 * e^(-x^2) as x approaches infinity
     
  4. Feb 24, 2008 #3
    well, i guess this is going to be zero, since [tex] e^{x^2}[/tex] grows much faster than x^3
     
  5. Feb 24, 2008 #4
    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.
     
  6. Feb 24, 2008 #5
    and if you want to use l'hopitals rule, you need to do it three times from what i can see!
     
  7. Feb 24, 2008 #6
    ah ok....I used l'hospital's rule 3 times and ended up with

    lim (3/(4x^3*e^(x^2)) = 0
     
  8. Feb 24, 2008 #7
    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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