Results 1 to 6 of 6

Math Help - Finding a limit

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    156

    Finding a limit

    \lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right).

    I worked out that it was an interminate difference of the form ∞ - ∞, so I figured I have to differentiate xlnx - x + 1 / (x-1)(lnx), but then the answer will be reaaally long, and I dont know what to do. This sometimes happens when I try to differentiate something
    Last edited by Chris L T521; May 11th 2010 at 05:18 AM. Reason: Formatted the question with LaTeX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by SyNtHeSiS View Post
    \lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right).

    I worked out that it was an interminate difference of the form ∞ - ∞, so I figured I have to differentiate xlnx - x + 1 / (x-1)(lnx), but then the answer will be reaaally long, and I dont know what to do. This sometimes happens when I try to differentiate something
    That is almost the correct way of going about it!

    \lim_{x\to1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)=\lim_{x\to 1}\frac{x\ln x - x+1}{(x-1)\ln x}

    When you evaluate this limit, we get the indeterminate case \frac{0}{0}. So to apply L'H˘pital's rule, we need to differentiate both numerator and denominator separately! Recall that L'H˘pital's rule says that if \lim_{x\to a}\frac{f(x)}{g(x)} is of the form \frac{\infty}{\infty} or \frac{0}{0} and if \lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)} exists, then \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}.

    So we don't look at \lim_{x\to 1}\frac{d}{dx}\left[\frac{x\ln x-x+1}{(x-1)\ln x}\right], we look at \lim_{x\to1}\frac{\dfrac{d}{dx}[x\ln x-x+1]}{\dfrac{d}{dx}[(x-1)\ln x]}!

    Does this clarify things? Can you finish this problem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    I have an interesting method but not a normal ...


     \frac{x}{x-1} - \frac{1}{\ln{x}}

     = \frac{ (x-1) + 1}{x-1} - \frac{1}{\ln{x}}

     = 1 + \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right )


    I let  L = \lim_{x\to 1} \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right )

    we know  \frac{1}{t^2 - 1} = \frac{1}{2} \frac{1}{t-1} - \frac{1}{2} \frac{1}{t+1} and so

     \frac{1}{x -1} = \frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1}

    so we have

     L = \lim_{x\to 1} \left(\frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1}  - \frac{1}{\ln{x}} \right )

    also  \ln{x} = 2\ln{\sqrt{x}}

     L = \lim_{x\to 1} \left( \frac{1}{2} ( \frac{1}{\sqrt{x} -1} - \frac{1}{ \ln{\sqrt{x}}}) \right) - \frac{1}{2} \lim_{x\to 1} \frac{1}{\sqrt{x} + 1}

    Let u = \sqrt{x}

     L = \frac{L}{2} - \frac{1}{4}

     L = -\frac{1}{2}

    so the final answer is  1 - \frac{1}{2} = \frac{1}{2}


    More surprisingly , if we continue to 'break down' the fraction  \frac{1}{ \sqrt{x} - 1} = \frac{1}{2} \frac{1}{\sqrt[4]{x} -1} - \frac{1}{2} \frac{1}{\sqrt[4]{x}+1}   and put  \ln{x} = 4\ln{\sqrt[4]{x}} and so on for  8,16,32 ,... . All of the situations can reach the answer obtained !!

    If there is no mistake in my method , it is interesting that the answer will not be affected for different choices of the base of the log. function !!

    EDIT:

    The answer will not be affected for different choices of the base of the log. function .
    This statement is false because at first we don't know if the limit exists , we have  L = \frac{L}{2} - \frac{1}{4} , but we are also forced to believe that  \infty = \frac{ \infty}{2} - \frac{1}{4} . The limit only exists when the base is  e .
    Last edited by simplependulum; May 11th 2010 at 07:21 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2010
    Posts
    156
    I tried differentiating and got:

    lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Quote Originally Posted by SyNtHeSiS View Post
    I tried differentiating and got:

    lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2
    \frac{d}{dx} \left[ x \ln x - x + 1 \right] = \frac{x}{x} + \ln x - 1 = \ln x

    \frac{d}{dx} \left[ (x-1) \ln x \right] = \frac{x-1}{x} + \ln x = 1 - \frac{1}{x} + \ln x

    So: \lim_{x \to 1} \, \frac{\ln x}{1 - \frac{1}{x} + \ln x}

    Again we get \frac{0}{0}, so we repeat the process.

    \frac{d}{dx} [\ln x] = \frac{1}{x}

    \frac{d}{dx} \left[ 1 - \frac{1}{x} + \ln x \right] = \frac{1}{x^2} + \frac{1}{x}

    So: \lim_{x \to 1} \, \frac{\frac{1}{x}}{\frac{1}{x^2} + \frac{1}{x}}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,389
    Thanks
    1324
    Quote Originally Posted by SyNtHeSiS View Post
    I tried differentiating and got:

    lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2
    No, that is NOT \infty, it is " \frac{0}{0}" so use L'Hopital's rule again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the limit
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 11th 2011, 07:50 AM
  2. Finding limit of this function, using Limit rules
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 27th 2011, 01:12 PM
  3. [SOLVED] finding limit
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 23rd 2011, 10:17 PM
  4. Finding a limit. Finding Maclaurin series.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2010, 10:04 PM
  5. Finding the limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 23rd 2009, 12:21 PM

Search Tags


/mathhelpforum @mathhelpforum