Oblique asymptote

**youngb11** · March 1st 2011, 01:39 PM

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.

**skeeter** · March 1st 2011, 02:08 PM

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

**topsquark** · March 1st 2011, 02:23 PM

Looks that way to me as well.

-Dan

**HallsofIvy** · March 2nd 2011, 02:41 AM

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, $y= \frac{2}{3}xj$ .

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