Results 1 to 2 of 2

Math Help - Horizontal Asymptote

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    54

    Horizontal Asymptote

    I need to find the horizontal asymptote for this function:

    y= x/((sqrt)x^2-1)

    the method we're supposed to use is multiple the numerator and denominator by (1/x). I just have some problems when square roots are involved.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    As long as you don't have to be super formal in your method, here's a way to do it.

    \sqrt{x^2-1}=\sqrt{x^2(1-\frac{1}{x^2})}=\sqrt{x^2}*\sqrt{(1-\frac{1}{x^2})}=x*\sqrt{(1-\frac{1}{x^2})}

    Now you can cancel out your x's and evaluate the new limit as x approaches infinity of \frac{1}{\sqrt{1-\frac{1}{x^2}}}, which should be trivial.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. horizontal asymptote
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 4th 2009, 10:09 AM
  2. Horizontal Asymptote
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 17th 2008, 02:30 PM
  3. horizontal asymptote
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 15th 2008, 01:51 PM
  4. Horizontal asymptote
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 23rd 2007, 12:13 PM
  5. Horizontal asymptote
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 16th 2007, 08:48 AM

Search Tags


/mathhelpforum @mathhelpforum