Results 1 to 7 of 7

Math Help - l'hospitals rule

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    37

    l'hospitals rule

    I have a problem xe^(1/x) and i'm supposed to use l'hospital's rule to explain its behavior as x -> 0. But L'hospital's rule is for division only I thought? Any help would be great thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by UMStudent View Post
    I have a problem xe^(1/x) and i'm supposed to use l'hospital's rule to explain its behavior as x -> 0. But L'hospital's rule is for division only I thought? Any help would be great thanks.
    we spoke about a problem similar to this just tonight. I'll find the post for you. but the short answer is:

    note that we can write x as 1/(1/x)

    so xe^(1/x) = 1/(1/x) * e^(1/x) = [e^(1/x)]/(1/x) ........this is a quotient, however, this goes to 1/infinity, which is not a condition to use l'hopital's, i think you left out an x somewhere or something

    see http://www.mathhelpforum.com/math-he...e-inf-inf.html
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2007
    Posts
    37
    xe^(1/x) is the exact problem thats why I was having trouble and the problem says to specifically use l'hospital's rule
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by UMStudent View Post
    xe^(1/x) is the exact problem thats why I was having trouble and the problem says to specifically use l'hospital's rule
    ah yes, you can use l'hopital's on it. the lim is x-->0, i thought it was x-->infinity since that's what the problem we were doing was like. you can follow my original guidelines then. l'hopital's will work
    Last edited by Jhevon; April 12th 2007 at 09:37 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    lim{x-->oo}xe^(1/x)
    = lim{x-->oo}(1/(1/x))e^(1/x)
    = lim{x-->oo} [e^(1/x)]/(1/x) .......this goes to inf/inf as x-->0, we can use l'hopital's

    Apply L'hopital's

    => lim{x-->oo}[e^(1/x)]/(1/x) = lim{x-->oo}[(-x^-2)e^(1/x)]/(-x^-2)
    = lim{x-->oo} e^(1/x)
    = infinity
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    UMStudent here is a good website I discovered earlier today. It goes through all of the possible scenarios. Just keep working with it and you will pick it up rather quickly.

    http://spot.pcc.edu/~sperry/M252_S-4_5.pdf
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2007
    Posts
    37
    Lol why didn't you say you were a genius. But thanks for all the help from both of you appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. l'hospitals rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 11th 2010, 05:09 AM
  2. l'hospitals rule help
    Posted in the Calculus Forum
    Replies: 8
    Last Post: November 16th 2009, 09:48 AM
  3. L'Hospitals rule and Mean Value Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 19th 2009, 07:41 PM
  4. L'hospitals rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2008, 02:21 PM
  5. L'Hospitals Rule
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 21st 2007, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum