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 problem with your second approach is that you've kept more precision than your approximation actually has — you can say $y \approx x$, but you don't have enough precision to clarify that more specifically to $y \approx x+1$ (or any other translate).
In more detail,
$$ \frac{1}{1 + \frac{2}{x}} \approx 1 - \frac{2}{x} $$
and consequently,
$$ \frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}}
\approx
\left( x+1-\frac{6}{x} \right) \left(1 - \frac{2}{x} \right) 
\approx
x \cdot 1 + 1 \cdot 1 - x \cdot \frac{2}{x}$$
By neglecting the $\frac{2}{x}$ term of the denominator, you neglect the $x \cdot \frac{2}{x}$ term of this approximation — but that term is $-2$, so you're neglecting a nonnegligible quantity!
Keeping the $\frac{2}{x}$ term around, the above approximation gives $x-1$, as desired.
For more rigor, you can use big O notation:
$$\frac{1}{1 + \frac{2}{x}} = 1 - \frac{2}{x} + O(x^{-2}) $$
$$ 
\frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}}
= \left( x+1+O(x^{-1}) \right) \left(1 - \frac{2}{x} + O(x^{-2}) \right)
= x - 1 + O(x^{-1})
$$
