Results 1 to 5 of 5

Math Help - limits of horizontal asymptotes

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    2

    limits of horizontal asymptotes

    How do you figure out the:

    lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2008
    Posts
    31
    Quote Originally Posted by used9090 View Post
    How do you figure out the:

    lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)
    You could divide all of the terms in the expression by the highest term in the denominator. The highest term in the denominator would be x^6, and the highest term in the numerator would be x^{11}. This way you can find out that there will be a x^5 expression left meaning the limit as x approaches infinity is infinity.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by used9090 View Post
    How do you figure out the:

    lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)
    Under standard order of operations, this would be \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3(4-x)}{(x-5)^5}, but I am guessing that you intended to write \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}\text.

    We have

    \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}

    =\lim_{x\to\infty}\frac{x^8(1-3/x)^8\left[x^3(2/x-1)^3\right]}{x^5(1-5/x)^5\left[x(4/x-1)\right]}

    =\lim_{x\to\infty}\frac{x^{11}}{x^6}=\lim_{x\to\in  fty}x^5=\infty
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    Posts
    2
    My teachers solution guide listed the answer to that specific problem as
    -OO though. Is there an error perhaps in my teachers arithmetic?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    31
    Well if the problem is the limit as that function approaches infinity, then yes. But if the problem is the limit as that function approaches negative infinity, then your teacher is correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Horizontal asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: July 13th 2010, 04:51 AM
  2. Horizontal asymptotes
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 2nd 2010, 09:25 AM
  3. Horizontal asymptotes/limits
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 28th 2009, 09:14 PM
  4. horizontal asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 17th 2008, 06:08 AM
  5. Replies: 2
    Last Post: June 23rd 2008, 11:04 PM

Search Tags


/mathhelpforum @mathhelpforum