Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.

I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:
$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.

The trouble is that the numerator in
$$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$
stays large, so a small error in the denominator is magnified. Better to write
$$ \frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots  $$
and multiply; the infinite series indicated converges for $|x| > 2.$
$$ \left( x+1-\frac{6}{x} \right) \left( 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3}  + \frac{16}{x^4} + \cdots \right) = x - 1 -  \frac{4}{x} +   \frac{8}{x^2} + \cdots $$
