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Math Help - asymptotes of ln

  1. #1
    ywd
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    asymptotes of ln

    Find all vertical and horizontal asymptotes of

    f(x) = ln(2x)/ln(x^2)

    I understand for vertical asymptote, there is one at x = 1, because denominator becomes 0.

    But for horizontal asymptote, i know as x -> inf, the denominator will increase faster than the numerator because of the squared, but is there a way to show it mathematically?

    Like with polynomials, you divide by the highest power, but with ln, what do you do?

    Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ywd View Post
    Find all vertical and horizontal asymptotes of

    f(x) = ln(2x)/ln(x^2)

    I understand for vertical asymptote, there is one at x = 1, because denominator becomes 0.

    But for horizontal asymptote, i know as x -> inf, the denominator will increase faster than the numerator because of the squared, but is there a way to show it mathematically?

    Like with polynomials, you divide by the highest power, but with ln, what do you do?

    Thanks.
    note that \ln 2x = \ln 2 + \ln x

    and

    \ln (x^2) = 2 \ln |x|
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  3. #3
    ywd
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    oh i see

    so its limit of

    (ln 2 + ln x)/(2ln x)

    => ln2/2lnx + lnx/2lnx

    => 0 + 1/2

    => 1/2

    :$ can you tell me if i did it right? we didn't cover this in our lecture >.<
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ywd View Post
    oh i see

    so its limit of

    (ln 2 + ln x)/(2ln x)

    => ln2/2lnx + lnx/2lnx

    => 0 + 1/2

    => 1/2

    :$ can you tell me if i did it right? we didn't cover this in our lecture >.<
    yes.

    we only had to take the limit going to infinity here, since the limit going to -infinity doesn't exist. thus, ln(|x|) = ln(x) for positive values, and so, the ln(x)'s cancelled, leaving you with what you have
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