A trigonometric function like f(x)=sin(x)f(x)=sin(x) is defined and continuous for all real numbers xx. Therefore, if aa 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.
