How to solve it please

lim ............................ (square root(cosx))
............... _............... ------------------------------
x-->(pi / 2) ........................ x - (pi/2)
thank you so much :)

(square root(cos (pi/2) ) = (square root(0) = 0

lim = 0/0

heard of l'Hospital ?

i.e. differentiate (only in these cases of indeterminism 0/0 and infinity/infinity) nominator + denominator SEPARATELY up to the point when the limit of the fraction is computable

your case: lim = 0/0 , hence start differentiating

nominator'= (square root(cosx)) ' = sin(x)/(2*square root(cosx))

denominator'= (x - pi/2 ) ' = 1

now limit = - [sin(x)/(2*square root(cosx))] / 1 = - sin(x)/[(2*square root(cosx))] = - 1 / (2 * 0 ) = - infinity

limit(sqrt(cos(x))/(x - (1/2)*Pi), x = (1/2)*Pi) = undefined