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Thread: In application of limits (Asymptotes)

  1. #1
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    In application of limits (Asymptotes)

    Hi

    In application of limits (Asymptotes)

    $\displaystyle f(x)=x+1/x$ is the function

    Hoxizontal Asymptotes

    solution

    $\displaystyle lim x-infinity f(x) = lim x-infinity [ x + 1/x ] = infinity $


    but Vertical Asymptotes why? $\displaystyle limx-0$

    $\displaystyle lim x-0 f(x) = lim x-0 [ x + 1/x ] = infinity $

    thanks a lot
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  2. #2
    Super Member Aryth's Avatar
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    You should graph this function, you will see that it is supposed to have a vertical asymptote at y = 0 and a horizontal one at x = 1 . If you look at the infinite limit you can split it up using the common denominator:

    $\displaystyle \frac{x + 1}{x} = 1 + \frac{1}{x}$

    This is equivalent to the function you had before. Take the limits at infinity and zero of that and you will get a different answer for the horizontal asymptote.
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