A trigonometric function like #f(x)=sin(x)# is defined and continuous for all real numbers #x#. Therefore, if #a# is a specific real number,

#lim_{x->a+}f(x)=lim_{x->a-}f(x)=lim_{x->a}f(x)=f(a)#.

For example, if #a=pi/3#, then

#lim_{x->pi/3+}f(x)=lim_{x->pi/3-}f(x)=lim_{x->pi/3}f(x)=sin(pi/3)=sqrt(3)/2#,

If #f(x)=csc(x)=1/(sin(x))#, then this is defined and continuous for all real numbers #x# except for integer multiples of #pi# (#0, \pm pi, \pm 2pi, \pm 3pi,\ldots#). And, for example, if #a=pi/3#:

#lim_{x->pi/3+}csc(x)=lim_{x->pi/3-}csc(x)=lim_{x->pi/3}csc(x)#

#=csc(pi/3)=2/sqrt(3)=(2sqrt(3))/3#.

On the other hand, the inverse trigonometric function #f(x)=sin^{-1}(x)=arcsin(x)# has a domain equal to the closed interval #[-1,1]={x:-1\leq x\leq 1}#. In this case, we are only "allowed" to let #x# approach #-1# from the right and #1# from the left. In these cases, we still end up evaluating the limit by substitution (the graph is "one-sided continuous" in those cases):

#lim_{x->-1+}sin^{-1}(x)=sin^{-1}(-1)=-pi/2# and

#lim_{x->1-}sin^{-1}(x)=sin^{-1}(1)=pi/2#

Another common set of one-sided limits to know are those for #cos^{-1}(x)=arccos(x)#:

#lim_{x->-1+}cos^{-1}(x)=cos^{-1}(-1)=pi# and

#lim_{x->1-}cos^{-1}(x)=cos^{-1}(1)=0#

For #tan^{-1}(x)=arctan(x)#, you have to let #x# approach #\pm \infty# "from one-side" (though you don't bother putting this in the notation). Technically, we are not doing any "substituting" of #\infty# here. We are just using knowledge about how #tan^{-1}(x)# is defined and how the vertical asymptotes of #tan(x)# correspond to horizontal asymptotes of #tan^{-1}(x)#.

#lim_{x->\infty}tan^{-1}(x)=pi/2# and #lim_{x->-\infty}tan^{-1}(x)=-\pi/2#.