Results 1 to 7 of 7

Math Help - Horizontal asymptote for e function

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    56

    Horizontal asymptote for e function

    Hello,
    The function is
    y = \frac{e^x + e^{-x}}{e^x - e^{-x}}

    I managed to get to the first horizontal asymptote by using lhospital's rule after I got to this:
    y = \frac{e^{2x} + 1}{e^{2x} - 1} = \frac{(e^{2x} + 1)'}{(e^{2x} - 1)'} = \frac{e^{2x} * 2}{e^{2x} * 2} = \frac{1}{1} = 1

    But then the sign of infinity which I insert doesn't matter because y will always be 1, and I know that there's another asymptote in y=-1.

    How do I get to the other asymptote? Have I done anything wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Horizontal asymptote for e function

    I would rewrite the function as:

    a) For the asymptote as x\to+\infty

    y=\frac{1+e^{-2x}}{1-e^{-2x}}

    \lim_{x\to+\infty}y=\frac{1+0}{1-0}=1

    b) For the asymptote as x\to-\infty

    y=\frac{e^{2x}+1}{e^{2x}-1}

    \lim_{x\to-\infty}y=\frac{0+1}{0-1}=-1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    56

    Re: Horizontal asymptote for e function

    How did you get to that function?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    fkf
    fkf is offline
    Junior Member fkf's Avatar
    Joined
    Nov 2012
    From
    Sweden
    Posts
    70
    Thanks
    9

    Re: Horizontal asymptote for e function

    If you're asking about a) then MarkFL2 is dividing both the numerator and denominator with e^x. Doing this wont change the value of the fraction and hence is a good way to get a fraction were we easy can calculate the limit.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Horizontal asymptote for e function

    Yes, for:

    a) \frac{e^x+e^{-x}}{e^x-e^{-x}}\cdot\frac{e^{-x}}{e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}

    b) \frac{e^x+e^{-x}}{e^x-e^{-x}}\cdot\frac{e^{x}}{e^{x}}=\frac{e^{2x}+1}{e^{2x}-1}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2009
    Posts
    56

    Re: Horizontal asymptote for e function

    So I always have to get to a certain equation of the same function in order to get its limits?
    How do I know when to keep trying to look for a way I can get the limit and when to let it go because it has no horizontal asymptote?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Horizontal asymptote for e function

    Sometimes it is helpful to rewrite a function to more easily determine its limit, particularly limits at plus/minus infinity. When there is no horizontal asymptote, you should be able to show that the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Horizontal asymptote of this function
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 9th 2011, 09:10 PM
  2. horizontal asymptote
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 4th 2009, 11:09 AM
  3. Replies: 2
    Last Post: March 26th 2009, 09:27 PM
  4. Replies: 1
    Last Post: November 25th 2008, 01:30 PM
  5. Horizontal asymptote
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 23rd 2007, 01:13 PM

Search Tags


/mathhelpforum @mathhelpforum