Results 1 to 6 of 6
Like Tree3Thanks
  • 1 Post By Cesc1
  • 1 Post By Soroban
  • 1 Post By FelixFelicis28

Math Help - Help Finding a Limit

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    USA
    Posts
    14

    Help Finding a Limit

    lim as x --> 1 of
    (x^(1/3) - 1) / (x^(1/2) - 1)

    The answer is supposed to be 2/3.

    I tried to use conjugates, but I still end up with a 0 in the denominator.

    Could someone work through it? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    826
    Thanks
    209

    Re: Help Finding a Limit

    Help Finding a Limit-20-sep-13-2.png
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2012
    From
    Mexico
    Posts
    53
    Thanks
    7

    Re: Help Finding a Limit

    Notice that you can express that as

    lim as x--->1 of
    f(x)/g(x)

    where f(x)= x^(1/3)-1, and g(x)=x^(1/2)-1

    Since both limits approach to 0, you can use L'Hopital's rule:

    lim as x---> c of

    f(x)/g(x)= lim as x-->c of

    f'(x)/g'(x)
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617

    Re: Help Finding a Limit

    Hello, IdentityProblem!

    \lim_{x\to1}\frac{x^{\frac{1}{3}} - 1}{x^{\frac{1}{2}} - 1}

    The conjugate of (x^{\frac{1}{2}}-1) is (x^{\frac{1}{2}}+1)

    The conjugate of (x^{\frac{1}{3}}-1) is (x^{\frac{2}{3}}+x^{\frac{1}{3}} + 1)

    Now try it again . . .
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2013
    From
    USA
    Posts
    14

    Re: Help Finding a Limit

    Thank you, Soroban!

    That raises a new question: how do you find a conjugate from scratch?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member FelixFelicis28's Avatar
    Joined
    Jan 2013
    From
    Edinburgh, United Kingdom
    Posts
    70
    Thanks
    32

    Re: Help Finding a Limit

    Quote Originally Posted by IdentityProblem View Post
    Thank you, Soroban!

    That raises a new question: how do you find a conjugate from scratch?
    I'm not exactly sure what you mean as a conjugate in this sense, but if you're asking how Soroban found the conjugate of x^{\frac{1}{3}} - 1 (what I'm going to presume is an expression y such that (x^{\frac{1}{3}} - 1) y = x - 1) then this is a useful trick to have in mind:

    x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1}) \\ \implies x - 1 = (x-y) (x^{n-1} + x^{n-2} y + \cdots + xy^{n-2} + y^{n-1}) \bigg|_{n = 3, \ x\mapsto x^{\frac{1}{3}}, \ y = 1
    Last edited by FelixFelicis28; September 21st 2013 at 02:09 PM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 28th 2013, 07:28 PM
  2. Finding limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 29th 2013, 09:02 AM
  3. Finding limit of this function, using Limit rules
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 27th 2011, 01:12 PM
  4. Finding a limit
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: August 1st 2010, 04:19 AM
  5. Finding a limit. Finding Maclaurin series.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2010, 10:04 PM

Search Tags


/mathhelpforum @mathhelpforum