Question

limit as x approaches pi/2 of (sinx)^2 / xcosx

Answer 1

Let x=(pi)/2+h, where h is small. cos(x)=cos((pi)/2)cos(h)-sin((pi)/2)sin(h)=-h when h is small. sin(x)=sin((pi)/2)cos(h)+cos((pi)/2)sin(h)=1 as h approaches zero. So the expression becomes 1/(-h) which goes to minus infinity as h approaches zero and x approaches (pi)/2.