Question

What are the asymptote(s) and hole(s), if any, of `f(x) =cos((pix)/2)/((x-1)(x+2))`?

Answer 1

`f(x)` has a hole at `x=1`, a vertical asymptote at `x=-2` and a horizontal asymptote `y=0` (according to modern usage).

Explanation:

Given:

`f(x) = cos((pix)/2)/((x-1)(x+2))`

Note that the denominator is zero when `x=1` or `x=-2`.

As `x->1`, then `cos((pix)/2) -> 0`. With both numerator and denominator tending to `0`, we may have a hole or we may have an asymptote.

Substituting `t = x-1`, we have:

`lim_(x->1) f(x) = lim_(t->0) cos((pit)/2+pi/2)/(t(t+3))`

`color(white)(lim_(x->1) f(x)) = lim_(t->0) -sin((pit)/2)/((pit)/2) * ((pi/2)/(t+3))`

`color(white)(lim_(x->1) f(x)) = -1 * pi/6`

`color(white)(lim_(x->1) f(x)) = -pi/6`

So `f(x)` has a hole at `x=1`

As `x->-2`, then `cos((pix)/2)->cos(-pi) = -1`

So the numerator is non-zero, while the denominator is zero. So there is a vertical asymptote at `x=-2`

For any real value of `x`, we have `abs(cos((pix)/2)) <= 1`, while `(x-1)(x+2) -> oo` as `x -> +-oo`.

Hence `f(x)` has a horizontal asymptote `y=0`.

Note that historically some people would not count this as an asymptote since the graph of `f(x)` crosses it infinitely many times, but it does actually tend towards the asymptote.

graph{cos((pix)/2)/((x-1)(x+2)) [-10.42, 9.58, -1.2, 1.2]}