Question

How do you find the asymptotes for `s(t)=t/(sin t)`?

Answer 1

`t=[(",180^\circ",",360^\circ",",\cdots,n180^\circ),(,",\pi",",2\pi",",\cdots,n\pi)]`

Explanation:

For there to be an asymptote, the denominator must equal `0`.

So, `sint=0`.

`arcsin(sin(t))=t=arcsin(0)`

`arcsin(0)=t=[(0^\circ),(0^c)]`

However, `sin(180^\circ)=sin(\pi)=0`

Also, `lim_(t->0)t/sin(t)=1`, and it can't be an asymptote.

`t=arcsin(0)=[(180^\circ,360^\circ,\cdots,n180^\circ),(\pi,2\pi,\cdots,n\pi)]`

Answer 2

`s(t)` has vertical asymptotes for `t = npi` where `n` is any non-zero integer.

It has a hole (removable singularity) at `t=0`.

It has no horizontal or slant asymptotes.

Explanation:

Given:

`s(t) = t/(sin t)`

Note that `s(t)` will be undefined whenever the denominator `sin t` is zero, that is when:

`t = npi" "` where `n` is any integer.

The numerator is always non-zero, except when `t=0`, so we have vertical asymptotes at all values:

`t = npi" "` where `n` is any non-zero integer.

When `t=0` both the numerator and denominator are `0`, so `s(t)` is undefined, so we need to look at behaviour as `t->0` to determine whether this is an asymptote or a hole (removable singularity).

Note that:

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

so it is possible to make `s(t)` continuous at `t=0` by redefining it:

`s_1(t) = { (1 " if " t = 0), (t / (sin t) " if " t != 0) :}`

That means that `(0,1)` is a removable singularity a.k.a. hole.

Note that `s(t)` has no horizontal or slant (oblique) asymptotes since it has vertical asymptotes for arbitrarily large values of `t`.

graph{(y-x/(sin x)) = 0 [-79.84, 80.16, -39.24, 40.76]}

graph{(y-x/(sin x))(x^2+(y-1)^2-0.002) = 0 [-2.335, 2.665, -0.49, 2.01]}