Dear Uncle Colin,

I’m working on finding horizontal asymptotes for rational functions. I normally do that by division, but my teacher wants me to do it by rearranging - and I don’t really know what’s going on there! Can you explain?

- Horizontal Asymptotes Leaving Me Outwitted Somehow

Hi, HALMOS, and thanks for your message!

Yes, horizontal asymptotes can be a bit tricky. Let’s recap two more simple methods before we look at your teacher’s - which I like, but wouldn’t necessarily recommend.

Suppose we have $y = \frac{4x}{x-1}$.

Piratical method

“Arr! What happens if $x=100$? $y=\frac{400}{99}$, a parrot’s whisker above 4. That can’t be a coincidence!”

I don’t recommend this as a way to solve such questions - it certainly wouldn’t get you full marks - but it’s a fair way to check your answer.


As you suggested in your message, your first instinct is to divide and see what comes out as the lead term.

For example, here you could divide top and bottom by $x$ to get $y=\frac{4}{1 - \frac{1}{x}}$. As $x$ gets large, $\frac{1}{x}$ becomes small and you’re left with a smidge more than 4.

This is perfectly valid, and if you’re comfortable with it, feel free to stick with it.


The rearrangement method tries to find $x$ in terms of $y$, and spot where it becomes undefined due to a zero denominator:

$(x+1)y=4x$ $y=x(4-y)$ $x=\frac{y}{4-y}$

This is clearly undefined when $y=4$, which is the equation of the horizontal asymptote.

What’s going on here? If you redrew the graph with the $x$- and $y$-axes flipped, you would find the horizontal asymptotes on the original graph were transformed into vertical asymptotes on the flipped graph. We get the new graph by rearranging so we have $x=$ (stuff), and those vertical asymptotes whenever (stuff) tries to divide by zero.

Hope that helps explain your teacher’s method - I don’t have a good argument for why that method is supposed to be better than one you have that works!

- Uncle Colin