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Thread: Finding an asymptote

  1. #1
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    Finding an asymptote

    how do you find the asymptotes of a curve like:

    $\displaystyle (x+3)/(x^2+4)$

    where the denominator is a power for eg.

    and why do you integrate $\displaystyle 3/x$ to get 3lnx and not $\displaystyle -9x^{-3}$??

    help very appreciated thanks!!
    Last edited by mr fantastic; May 13th 2010 at 07:43 PM. Reason: Fixed some latex
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  2. #2
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by Yehia View Post
    how do you find the asymptotes of a curve like:

    $\displaystyle (x+3)/(x^2+4)$

    where the denominator is a power for eg.

    and why do you integrate $\displaystyle 3/x$ to get 3lnx and not $\displaystyle -9x^-3$??

    help very appreciated thanks!!
    To get the asymptotes, test its end behaviors by taking both limits as $\displaystyle x\rightarrow \infty$ and as $\displaystyle x\rightarrow -\infty$. Both of which are 0, therefore there is only one asymptote which is $\displaystyle y = 0$.

    To answer your second question, the reason it turns into $\displaystyle y = ln |x| $ is because when you are integrating using the reverse power rule which says

    $\displaystyle \int {x}^{r}dx = \frac{{x}^{r+1}}{r+1} $

    however this only works if $\displaystyle r \neq -1$, because then denominator becomes 0. In your case, we have $\displaystyle y = 3{x}^{-1}$, so $\displaystyle r = -1$ so it doesn't work. So we use $\displaystyle y = ln |x| $. Make sure that it's absolute value of x and not just regular x. If you don't put the absolute value signs, you'll get it wrong.

    Cheers!
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Yehia View Post
    and why do you integrate $\displaystyle 3/x$ to get 3lnx and not $\displaystyle -9x^{-3}$?
    A better question is, why do you think that the integral of 3/x is $\displaystyle -9x^{-3}$?
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