Alright, in the back of the book it says the horizontal asymptote is $\displaystyle y = 2$. I know that's true because when I went and graphed it, the lines were scared of that number.

Anyway, here is the equation:

$\displaystyle f(x) = 2+ \frac{1}{x}$

$\displaystyle n < m$ makes $\displaystyle y= 0 $ is what I thought but there's a two.

No idea what to do here.

Wouldn't getting rid of the bottom make it:

$\displaystyle f(x) = 2+ \frac{1}{x}$

$\displaystyle f(x) = 2x+1$

Unless...

$\displaystyle \frac{2}{1}+\frac{1}{x}$

$\displaystyle \frac{2x}{x}+\frac{1}{x}$

$\displaystyle = \frac{2x+1}{x}$

which makes the horizontal asymptote 2...

Never saw an example period of that though.