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

##### 2 Answers

#### Answer:

#### Explanation:

For there to be an asymptote, the denominator must equal

So,

However,

Also,

#### Answer:

It has a hole (removable singularity) at

It has no horizontal or slant asymptotes.

#### Explanation:

Given:

#s(t) = t/(sin t)#

Note that

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

The numerator is always non-zero, except when

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

When

Note that:

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

so it is possible to make

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

That means that

Note that

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]}