How do you find the limit of #(1-sintheta)/(theta-pi/2)# as #theta->pi/2#?

Question

3017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-28T17:05:56

#0.#

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.!