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

Question

Elijah Allen · Accepted Answer

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

Sophie Adams · 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.

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]}

Madelyn Anderson · Answer

undefined To find the asymptotes for the function $ s(t) = \frac{t}{\sin(t)} $, we need to identify where the function approaches infinity or negative infinity.

Asymptotes can occur where the denominator of a fraction approaches zero. In this case, $\sin(t)$ approaches zero at values of $ t $ such that $ t = k\pi $, where $ k $ is an integer. Therefore, we need to find the vertical asymptotes at these points.

The vertical asymptotes occur at $ t = k\pi $ where $ k $ is an integer, because at these points, the function $ s(t) $ approaches either positive or negative infinity.

Additionally, we need to check for any horizontal asymptotes. For this, we need to examine the behavior of the function as $ t $ approaches positive or negative infinity.

As $ t $ approaches positive or negative infinity, $ \sin(t) $ oscillates between -1 and 1, causing the function $ s(t) $ to oscillate without approaching any finite value. Therefore, there are no horizontal asymptotes for $ s(t) = \frac{t}{\sin(t)} $.

Thus, the asymptotes for $ s(t) = \frac{t}{\sin(t)} $ are vertical lines at $ t = k\pi $, where $ k $ is an integer.