讲的详细点.lim(x->+∞) sin^-1(x/(1-2x))

=-pi/6.
Using the continuity of sin^-1(x) at x= - 1/2,you can exchange the operation order and write it as
lim(x->+∞) sin^-1(x/(1-2x)) =sin^-1[lim(x->+∞) (x/(1-2x))]
Use the L'Hopital's rule for the limit
lim(x->+∞) (x/(1-2x)) = lim(x->+∞) ((dx/dx)/(d(1-2x)/dx)) = - 1/2,and sin^-1(-1/2)= -pi/6
therefore lim(x->+∞) sin^-1(x/(1-2x))=-pi/6有不用l'hopital's rule的吗。。我还没学到那you can also divide every term in x/(1-2x) by x, x/(1-2x) = 1/(1/x-2), and lim(x->+∞) 1/(1/x-2) = 1/(0-2)=-1/2