Results 1 to 4 of 4

Math Help - problematic limit

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    54

    problematic limit

    hi!

    i have a problem in finding limit for this funciton f (x) = x*(lnx)^23,where x --> 0. i thought i will be able to use Lopital's rule, but cannot transformate funcition to uncertainity in order to use Lopital's rule. Do anyone have ideas? And I also think that there should be way to find limit without Lopital's rule, using formulas like e^x-1~x.
    But anyway, i have no ideas for the moment. Can anybody help?

    thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    From
    Mumbai
    Posts
    83
    As x tends to zero, the given expression becomes an indeterminate of the form

    So convert it into an indeterminate form which allows application of L'Hospital rule .

    Like this :


    Now consider the numerator, and the denominator

    Both terms become infinite as x tends to zero, so L'Hospital's rule is applicable.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    In fact you can apply L'hopital:

    x\cdot\ln(x)^{23} =  \frac{\ln(x)^{23}}{\frac{1}{x}} = [-\frac{\infty}{\infty}] (as x --> 0)

    by repeatedly applying L'hopital you should find: \lim_{x\to 0}x\cdot \ln(x)^{23} = \lim_{x\to 0}-23! x = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744
    Hello, waytogo!

    Ah, Dinkydoe beat me to it . . .
    Well, I'm not going to delete all this!


    \lim_{x\to0}x\!\cdot\!(\ln x)^{23}

    As given, the limit is: . 0\cdot(-\infty)


    We have: . \frac{(\ln x)^{23}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty} . .
    We can apply L'Hoptal's Rule

    Apply L'Hopital: . \frac{23\cdot (\ln x)^{22}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot(\ln x)^{22}}{x^{-1}}\;\rightarrow\;\frac{\text{-}\infty}{\infty} . .
    We can apply L'Hoptal's Rule.

    Apply L'Hopital: . -\frac{23\cdot22\cdot(\ln x)^{21}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23\cdot22\cdot(\ln x)^{21}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}

    Apply L'Hopital: . \frac{23\cdot22\cdot21\cdot(\ln x)^{20}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdot(\ln x)^{20}}{x^{-1}}  \;\rightarrow\; \frac{\text{-}\infty}{\infty}

    . . . . . \vdots

    Apply L'Hopital: . \frac{23\cdot22\cdot21\cdots 3\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdots3\cdot(\ln x)^2}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}

    Apply l"Hopital: . \frac{23\cdot22\cdot21\cdots2\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23!(\ln x)}{x^{-1}} \;\rightarrow\; \frac{\text{-}\infty}{\infty}

    Apply L'Hopital: . \frac{23!\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23!}{x^{-1}} \;=\;-23!x


    Therefore: . \lim_{x\to0}\,(-23!x) \;=\;0

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. problematic integral #2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 21st 2010, 03:54 AM
  2. problematic integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 20th 2010, 09:31 AM
  3. Problematic Inverses
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 11th 2009, 08:20 AM
  4. The sum of another problematic sequence.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 22nd 2008, 04:14 AM
  5. The sum of a most problematic sequence.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 20th 2008, 11:01 AM

Search Tags


/mathhelpforum @mathhelpforum