Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x) = x/ sinx`?

Answer 1

`f(x)` has vertical asymptotes at `x=n pi` for any non-zero integer `n`.

It has a hole at `x=0`

Explanation:

Vertical asymptotes will occur where the denominator is zero and the numerator non-zero.

`sin x = 0` if and only if `x = n pi` for some `n in ZZ`

Hence `f(x)` has vertical asymptotes at `x = n pi` where `n in ZZ` and `n != 0`

`f(x)` has a hole at `x=0`. The rational expression becomes `0/0`, which is undefined, but the right and left limits exist at `x=0` and are equal:

`lim_(x->0) x/(sin x) = 1`

`f(x)` has no horizontal or slant asymptotes, since we find:

`f(2n pi+pi/2) = 2n pi+pi/2 -> +oo` as `n -> +oo`

`f(2n pi+pi/2) = 2n pi+pi/2 -> -oo` as `n -> -oo`

`f(2n pi+(3pi)/2) = -(2n pi+(3pi)/2) -> -oo` as `n -> +oo`

`f(2n pi+(3pi)/2) = -(2n pi+(3pi)/2) -> +oo` as `n -> -oo`