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

2 Answers
Nov 9, 2017

Answer:

#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)]#

Nov 11, 2017

Answer:

#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]}