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

Nov 9, 2017

$t = \left[\left(\text{,180^\circ",",360^\circ",",\cdots,n180^\circ),(,",\pi",",2\pi",} , \setminus \cdots , n \setminus \pi\right)\right]$

#### Explanation:

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

So, $\sin t = 0$.

$\arcsin \left(\sin \left(t\right)\right) = t = \arcsin \left(0\right)$

$\arcsin \left(0\right) = t = \left[\begin{matrix}{0}^{\setminus} \circ \\ {0}^{c}\end{matrix}\right]$

However, $\sin \left({180}^{\setminus} \circ\right) = \sin \left(\setminus \pi\right) = 0$

Also, ${\lim}_{t \to 0} \frac{t}{\sin} \left(t\right) = 1$, and it can't be an asymptote.

$t = \arcsin \left(0\right) = \left[\begin{matrix}{180}^{\setminus} \circ & {360}^{\setminus} \circ & \setminus \cdots & n {180}^{\setminus} \circ \\ \setminus \pi & 2 \setminus \pi & \setminus \cdots & n \setminus \pi\end{matrix}\right]$

Nov 11, 2017

$s \left(t\right)$ has vertical asymptotes for $t = n \pi$ 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:

$s \left(t\right) = \frac{t}{\sin t}$

Note that $s \left(t\right)$ will be undefined whenever the denominator $\sin t$ is zero, that is when:

$t = n \pi \text{ }$ where $n$ is any integer.

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

$t = n \pi \text{ }$ where $n$ is any non-zero integer.

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

Note that:

${\lim}_{t \to 0} \frac{t}{\sin t} = 1$

so it is possible to make $s \left(t\right)$ continuous at $t = 0$ by redefining it:

${s}_{1} \left(t\right) = \left\{\begin{matrix}1 \text{ if " t = 0 \\ t / (sin t) " if } t \ne 0\end{matrix}\right.$

That means that $\left(0 , 1\right)$ is a removable singularity a.k.a. hole.

Note that $s \left(t\right)$ 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]}