Results 1 to 6 of 6
Like Tree3Thanks
  • 1 Post By HallsofIvy
  • 1 Post By romsek
  • 1 Post By Prove It

Math Help - limits question

  1. #1
    Junior Member
    Joined
    Mar 2014
    From
    United Kingdom
    Posts
    34

    limits question

    lim x-->0

    \frac{{x}^{x} -1}{x lnx}

    also it says assume as x--> 0 xlnx --> 0

    my answer is its 0/0 type since anything to the power of 0 is 1 including  {x}^{x} ?

    using L H Rule you get

    \frac{{x}^{x} (1+lnx)}{(1+lnx)}

    then again the anything to the power of zero thing gives you that {x}^{x} --> 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,419
    Thanks
    1330

    Re: limits question

    There is no such "thing". 0 to the 0 power is NOT 1. What L'Hopital's rule says is that the limit is the same as the limit of x^x. To find that limit, let y= x^x so that ln(y)= x ln(x). If you are allowed to "assume" the limit of that is 0, then the limit of ln(y) is 0 so the limit of x^x is, indeed, 1.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,238
    Thanks
    851

    Re: limits question

    Quote Originally Posted by Applestrudle View Post
    lim x-->0

    \frac{{x}^{x} -1}{x lnx}

    also it says assume as x--> 0 xlnx --> 0

    my answer is its 0/0 type since anything to the power of 0 is 1 including  {x}^{x} ?

    using L H Rule you get

    \frac{{x}^{x} (1+lnx)}{(1+lnx)}

    then again the anything to the power of zero thing gives you that {x}^{x} --> 1
    what you need to properly finish this is the fact that

    $\large x^x = e^{x \ln(x)}$

    you were told that $\displaystyle{\lim_{x \to 0}}~ x \ln(x) = 0$ so $\displaystyle{\lim_{x \to 0}}~e^{x \ln(x)} =e^0 = 1$
    Thanks from Applestrudle
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: limits question

    It's not that difficult to show \lim_{x\to 0} x\ln{x} = 0. If you write x\ln{x} = \frac{\ln{x}}{\frac{1}{x}}, then using L'Hopital's rule gives \frac{\frac{1}{x}}{-\frac{1}{x^2}}, which is -x.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: limits question

    Quote Originally Posted by hollywood View Post
    It's not that difficult to show \lim_{x\to 0} x\ln{x} = 0. If you write x\ln{x} = \frac{\ln{x}}{\frac{1}{x}}, then using L'Hopital's rule gives \frac{\frac{1}{x}}{-\frac{1}{x^2}}, which is -x.

    - Hollywood
    Actually the limit $\displaystyle \begin{align*} \lim_{x \to 0} x\ln{(x)} \end{align*}$ DOES NOT EXIST as $\displaystyle \begin{align*} x\ln{(x)} \end{align*}$ is only defined where $\displaystyle \begin{align*} x > 0 \end{align*}$. However, $\displaystyle \begin{align*} \lim_{x \to 0^+} x\ln{(x)} = 0 \end{align*}$.
    Thanks from hollywood
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: limits question

    Good point.

    Thanks,
    Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limits question
    Posted in the Calculus Forum
    Replies: 8
    Last Post: March 23rd 2013, 05:18 PM
  2. Limits question again
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2011, 06:12 PM
  3. Limits Question.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 1st 2010, 10:40 AM
  4. Limits question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 5th 2009, 05:35 AM
  5. Question on Limits
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 3rd 2008, 03:08 AM

Search Tags


/mathhelpforum @mathhelpforum