# Oblique asymptote

• Mar 1st 2011, 01:39 PM
youngb11
Oblique asymptote
Not sure if this is the correct section, but in the following function, $\displaystyle f(x)=\frac{2x^5-3x^2+5}{3x^4+5x-4}$, is the oblique asymptote $\displaystyle y=\frac{2}{3}x$? I ask because the answer in the book is different, but I can't see my mistake.
• Mar 1st 2011, 02:08 PM
skeeter
Quote:

Originally Posted by youngb11
Not sure if this is the correct section, but in the following function, $\displaystyle f(x)=\frac{2x^5-3x^2+5}{3x^4+5x-4}$, is the oblique asymptote $\displaystyle y=\frac{2}{3}x$? I ask because the answer in the book is different, but I can't see my mistake.

divide every term by $x^4$ ...

$f(x) = \dfrac{2x - \frac{3}{x^2} + \frac{5}{x^4}}{3 + \frac{5}{x^3} - \frac{4}{x^4}}$

looks like $y = \dfrac{2x}{3}$ to me
• Mar 1st 2011, 02:23 PM
topsquark
Looks that way to me as well.

-Dan
• Mar 2nd 2011, 02:41 AM
HallsofIvy
Even a "long division" gives
$\frac{2x^5- 3x^2+ 5}{3x^4+ 5x- 4}= \frac{2}{3}x- \frac{\frac{13}{3}x^2- \frac{8}{3}x- 5}{3x^4+ 5x- 4}$
so the oblique asymptote is, indeed, $y= \frac{2}{3}xj$.