Results 1 to 6 of 6
Like Tree5Thanks
  • 1 Post By romsek
  • 2 Post By HallsofIvy
  • 2 Post By Plato

Thread: limite expo 2

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    593
    Thanks
    3

    limite expo 2

    limite expo  2-gif.gif
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,314
    Thanks
    2709

    Re: limite expo 2

    2 rounds of applying L'Hopital's rule works.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,032
    Thanks
    3160

    Re: limite expo 2

    Setting x= -1 gives 0/0 so I would use L'Hopital's rule: The derivative of $\displaystyle (x+ 1)^2e^{x+ 1}$ is $\displaystyle 2(x+ 1)e^{x+ 1}+ (x+ 1)^2e^{x+ 1}= (x+ 1)(x+ 2)e^{x+ 1}$ and the derivative of $\displaystyle 1+ xe^{x+ 1}$ is $\displaystyle e^{x+ 1}+ xe^{x+ 1}= (x+ 1)e^{x+ 1}$. Those are both still 0 at x= -1 so do it again. The derivative $\displaystyle (x+ 1)(x+ 2)e^{x+ 1}$ is $\displaystyle (x+ 2)e^{x+ 1}+ (x+ 1)e^{x+ 1}+ (x+ 1)(x+ 2)e^{x+ 1}= (x+ 2+ x+ x^2+ 3x+ 2)e^{x+1}= (x^2+ 5x+ 4)e^{x+1}$ and the derivative of $\displaystyle (x+ 1)e^{x+ 1}$ is $\displaystyle e^{x+ 1}+ (x+ 1)e^{x+ 1}= (x+ 2)e^{x+ 1}$. At x= -1 those are 0 and 1. The limit is 0/1= 0.
    Thanks from topsquark and SlipEternal
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,072
    Thanks
    2973
    Awards
    1

    Re: limite expo 2

    Quote Originally Posted by dhiab View Post
    Click image for larger version. 

Name:	gif.gif 
Views:	1 
Size:	1.8 KB 
ID:	39037
    Use L'H˘pital's rule.

    $\displaystyle \mathop {\lim }\limits_{x \to - 1} \frac{{{{(x + 1)}^2}{e^{x + 1}}}}{{1 + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1){e^{x + 1}} + {{(x + 1)}^2}{e^{x + 1}}}}{{{e^{e + x}} + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1) + {{(x + 1)}^2}}}{{x + 1}} = ~?$
    Thanks from SlipEternal and topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2016
    From
    Earth
    Posts
    214
    Thanks
    97

    Re: limite expo 2

    Quote Originally Posted by HallsofIvy View Post
    / . . . / At x= -1 those are 0 and 1. The limit is 0/1= 0.
    I just entered this at WolframAlpha, for x between -2 to 0. The limit looks to be 2 instead of 0.


    Here:

    https://www.wolframalpha.com/input/?...+-2+to+x+%3D+0

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    Quote Originally Posted by Plato View Post
    Use L'H˘pital's rule.

    $\displaystyle \mathop {\lim }\limits_{x \to - 1} \frac{{{{(x + 1)}^2}{e^{x + 1}}}}{{1 + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1){e^{x + 1}} + {{(x + 1)}^2}{e^{x + 1}}}}{{{e^{e + x}} + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1) + {{(x + 1)}^2}}}{{x + 1}} = ~?$

    I thought something looked off. Those exponents in the denominator are not supposed to be "e + x."
    They are supposed to be "x + 1" just as in the numerators:


    $\displaystyle \mathop {\lim }\limits_{x \to - 1} \frac{{{{(x + 1)}^2}{e^{x + 1}}}}{{1 + x{e^{x + 1}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1){e^{x + 1}} + {{(x + 1)}^2}{e^{x + 1}}}}{{{e^{x + 1}} + x{e^{x + 1}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1) + {{(x + 1)}^2}}}{{x + 1}} = ~?$


    And then Plato's approach shows that, with one application of L'H˘pital's rule that the answer, which looks consistent with WolframAlpha, equals 2.
    Last edited by greg1313; Oct 24th 2018 at 10:02 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    593
    Thanks
    3

    Re: limite expo 2

    Hello .. here is a solution :limite expo  2-sans_t20.png
    let : x+1 = y
    Last edited by dhiab; Oct 24th 2018 at 02:20 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limite expo - sinus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 17th 2018, 03:03 AM
  2. Expo-equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 11th 2014, 06:07 AM
  3. Expo-equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 29th 2010, 11:46 PM
  4. logs and expo
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 21st 2009, 12:28 PM
  5. Limite
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Aug 24th 2009, 02:13 PM

/mathhelpforum @mathhelpforum