Results 1 to 4 of 4

Math Help - limit as x-> infinity

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    96

    limit as x-> infinity

    okay, so the question is what is the limit as x-> infinity of (sqrt(x^2+x))-(sqrt(x^2-x))?
    I know it is 1, but how do i show work / come to that conclusion?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    What about if you multiply the limit by \frac{\sqrt {x^2 + x} + \sqrt {x^2 - x}}{\sqrt {x^2 + x} + \sqrt {x^2 - x}}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Quote Originally Posted by mistykz View Post
    okay, so the question is what is the limit as x-> infinity of (sqrt(x^2+x))-(sqrt(x^2-x))?
    I know it is 1, but how do i show work / come to that conclusion?
    A nice trick, probably contrary to what has been beaten into your brain since Algebra I. UNrationalize it!!

    In other words, multiply by \frac{\sqrt{x^{2}+x}+\sqrt{x^{2}-x}}{\sqrt{x^{2}+x}+\sqrt{x^{2}-x}}.

    Resolve the numerator and wonderful things should appear. At least there should be somewhere to go with it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \displaystyle\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{  x^2-x}}=
    \displaystyle =\lim_{x\to\infty}\frac{2x}{x\left(\sqrt{1+\frac{1  }{x}}+\sqrt{1-\frac{1}{x}}\right)}=\lim_{x\to\infty}\frac{2}{\sq  rt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limit as x -> infinity?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 28th 2011, 05:22 PM
  2. Replies: 6
    Last Post: April 23rd 2011, 12:05 PM
  3. Replies: 3
    Last Post: February 1st 2011, 12:36 PM
  4. Limit at infinity equaling infinity?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 1st 2010, 10:07 PM
  5. Limit as n tends to infinity of...
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 2nd 2009, 04:06 PM

Search Tags


/mathhelpforum @mathhelpforum