<s>[INST] 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.
 [/INST] $\frac{y}{x+1} \;\to\; 1$ indeed, but:
$$y - (x+1) = \frac{x^2 + x - 6 - x^2 - 3 x - 3}{x+2} = \frac{-2 x - 9}{x+2} \quad \to \quad -2$$
which is why the slope of $x+1$ is correct, but its intercept is off by $-2$.
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