Results 1 to 6 of 6

Thread: Slant Asymptote

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    4

    Slant Asymptote

    Given the function f(x)=\frac{3x^4+4}{x^3-3x} I have found, by long division, that there is a slant asymptote, y=3x.
    Further, I wanted to find the same result by the classic limit definition.

    \lim_{x \mapsto \infty}\left [ \frac{3x^4+4}{x^3-3x} \right ]

    \lim_{x \mapsto \infty}\left [ \frac{x^4(1+\frac{4}{x^4})}{x^4(\frac{1}{x}-\frac{3}{x^3})} \right ]

    \lim_{x \mapsto \infty}\left [ \frac{1+\frac{4}{x^4}}{\frac{1}{x}-\frac{3}{x^3}} \right ]=\frac{1+0}{0-0}

    I ended up in an indeterminate form.The result should had be the same!What's wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    15,079
    Thanks
    3175

    Re: Slant Asymptote

    \frac{1}{0} is not an indeterminate form.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2011
    Posts
    4

    Re: Slant Asymptote

    I made a mistake in the simplification.The final result is \frac{3}{0}. The problem is division by zero!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    15,079
    Thanks
    3175

    Re: Slant Asymptote

    Quote Originally Posted by joaofonseca View Post
    I made a mistake in the simplification.The final result is \frac{3}{0}. The problem is division by zero!
    which means the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2011
    Posts
    4

    Re: Slant Asymptote

    The limit exist!Look at the graph:

    Slant Asymptote-slant.jpg
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    15,079
    Thanks
    3175

    Re: Slant Asymptote

    I'm afraid you are confused ... the limit as x \to \infty will determine a horizontal asymptote, but not a slant asymptote.

    After you do the long division, you get an expression of the form y = mx + b + r(x), where r(x) is a remainder that tends to 0 as x \to \infty , telling you that the function value approaches the line y = mx+b
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. slant asymptote
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jan 24th 2016, 06:34 PM
  2. Slant asymptote
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 14th 2009, 08:41 AM
  3. Slant asymptote
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 28th 2009, 09:33 PM
  4. Slant asymptote
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: Oct 21st 2008, 12:53 PM
  5. Slant Asymptote
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 20th 2008, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum