Recall that, #lim_(x to a) (f(x)-f(a))/(x-a)=f'(a).#
So, we have, #lim_(theta to pi/2) (1-sintheta)/(theta-pi/2)#
#=-lim_(theta to pi/2) {sintheta-sin(pi/2)}/(theta -pi/2)#
#=-sin'(pi/2)#
#=-cos(pi/2)#
#=0.#
In the following Direct Method to find the limit, we will use the
Standard Limit : #lim_(x to 0) sinx/x=1.#
Let, #l=lim_(theta to pi/2) (1-sintheta)/(theta-pi/2).#
Subst. #theta=h+pi/2," so that, as "theta to pi/2, h to 0.#
#:. l=lim_(h to 0) (1-sin(pi/2+h))/h,#
#=lim (1-cos h)/h=lim {(1-cos h)/h}{(1+cos h)/(1+cos h)}#
#=lim sin^2 h/{h(1+cos h)}#
#=lim_(h to 0) {sin h/h}{sinh /(1+cos h)}#
#=(1){0/(1+1)}#
#=0.#
Enjoy Maths.!
