# Asymptotes of a function

• Apr 13th 2009, 04:10 AM
Coach
Asymptotes of a function
Hello everyone

I am asked to find the asymptotes of $f(x)=\frac{x^2+1}{x+2}$

One of these is x=-2 , but when I try to arrive at the other asymptote, I get differing answers depending on the method I use.

First I tried to simplify the function

$
\frac{x^2 (1-\frac{1}{x^2})}{x(1+\frac{2}{x})}$

Looking at this it seems that when x approaches infinity, the function f(x) approaches the asymptote y=x.

However, if I use long division of polynomials of nested synthetic division I always get that the function will approach y=x-2.
• Apr 13th 2009, 05:13 AM
metlx
$(x^2 + 1) : (x + 2) = x -2 + \frac{5}{x + 2}$
$-(x^2 + 2x)$
-------------
$-2x+1$
$-(-2x - 4)$
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$5$

yeah it is y = x -2

I can't think of a way to do this with limits, since both x and y go to +- infinity.
• Apr 13th 2009, 10:44 AM
Coach
So why did the simplification yield a different result?