Question

What are the asymptote(s) and hole(s), if any, of `f(x) = x/(x-1)-(x-1)/x`?

Answer 1

`x=0` is an asymptote.

`x=1` is an asymptote.

Explanation:

First, let's simplify this so that we have a single fraction that we can take the limit of.

`f(x) = (x(x))/((x-1)(x)) - ((x-1)(x-1))/(x(x-1))`

`f(x) = (x^2 - (x-1)^2)/((x-1)(x)) = (x^2 - (x^2 - 2x + 1))/((x-1)(x))`

`f(x) = (2x-1)/((x-1)(x))`

Now, we need to check for discontinuities. This is just anything that will make the denominator of this fraction `0`. In this case, to make the denominator `0`, `x` could be `0` or `1`. So let's take the limit of `f(x)` at those two values.

`lim_(x->0)(2x-1)/(x(x-1)) = (-1)/(-1*0) = +-oo`

`lim_(x->1)(2x-1)/(x(x-1)) = 3/(1*0) = +-oo`

Since both of these limits tend towards infinity, both `x=0` and `x=1` are asymptotes of the function. There are therefore no holes in the function.