Finding an asymptote

Nov 2009
60
1
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 a moderator:
Apr 2010
70
23
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!
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
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 \(\displaystyle -9x^{-3}\)?