Results 1 to 5 of 5

Math Help - Another limit (known)

  1. #1
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13

    Another limit (known)

    Prove that \lim_{x\to0}\frac{\ln(1+x)}x=1

    Of course, without L'H˘pital's Rule
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    We can use the series expansion of ln(1+x)

    ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-............

    Dividing by x gives:

    1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-..........

    Of course, we can see if x approaches 0, we are left with 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Okay

    Now, without series?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, Krizalid!

    Recall the definitions of e.

    . . e \;\;=\;\;\lim_{x\to0}(1 + x)^{\frac{1}{x}} \;=\;\lim_{z\to\infty}\left(1 + \frac{1}{z}\right)^z


    Prove that: . \lim_{x\to0}\frac{\ln(1+x)}x=1

    Of course, without L'H˘pital's Rule

    \text{We have: }\;\frac{\ln(1+x)}{x} \;\;=\;\;\frac{1}{x}\!\cdot\!\ln(1+x) \;\;=\;\;\ln(1+x)^{\frac{1}{x}}

    \text{Then: }\;\lim_{x\to0}\left[\ln(1+x)^{\frac{1}{x}}\right] \;\;=\;\;\ln\underbrace{\left[\lim_{x\to0}(1+x)^{\frac{1}{x}} \right] }_{\text{This is }e} \;\;=\;\;\ln e \;\;=\;\;1

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Yeah

    I had exactly the same proof

    Cheers,
    K.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: August 26th 2010, 10:59 AM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  4. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  5. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum