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Math Help - Asymptotes of a function

  1. #1
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    Asymptotes of a function

    y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?
    Last edited by mr fantastic; September 20th 2009 at 01:13 AM. Reason: Moved from another thread.
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  2. #2
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    Quote Originally Posted by VitaX View Post
    y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?
    You cannot substitute infinity. It's not a number.

    Approach 1:

    y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}.

    \lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2.

    Therefore \lim_{x \rightarrow \pm \infty} y = 2 and so y = 2 is the horizontal asymptote.


    Approach 2:

    y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}

    and by inspection the horizontal asymptote is y = 2.


    Other approaches are possible.
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    Member VitaX's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You cannot substitute infinity. It's not a number.

    Approach 1:

    y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}.

    \lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2.

    Therefore \lim_{x \rightarrow \pm \infty} y = 2 and so y = 2 is the horizontal asymptote.


    Approach 2:

    y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}

    and by inspection the horizontal asymptote is y = 2.


    Other approaches are possible.
    You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.
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    Quote Originally Posted by VitaX View Post
    You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.
    No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.

    When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, \lim_{x \rightarrow + \infty} \frac{6}{x} = 0 is the correct way of treating it.
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    Quote Originally Posted by mr fantastic View Post
    No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.

    When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, \lim_{x \rightarrow + \infty} \frac{6}{x} = 0 is the correct way of treating it.
    Yes i understand I just didn't say take the limit of 1/x as x approaches infinity. I understand the terminology, just not familiar with latex.
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