Is that infinite ? #lim x-> pi/2 = ((x-pi/2))/((1-sinx)cosx) # Calculus Differentiating Trigonometric Functions Limits Involving Trigonometric Functions 1 Answer Cesareo R. Mar 1, 2017 #-oo# Explanation: Calling #y=x-pi/2# we have # ((x-pi/2))/((1-sinx)cosx) = -y/((1-cos(y))sin(y))# and also #lim_( x-> pi/2) ((x-pi/2))/((1-sinx)cosx) =lim_(y->0) -y/((1-cos(y))sin(y))=# #= lim_(y->0)1/(1-cos(y))lim_(y->0)(-y/siny) = oo (-1) = -oo# Answer link Related questions How do you find the limit of inverse trig functions? How do you find limits involving trigonometric functions and infinity? What is the limit #lim_(x->0)sin(x)/x#? What is the limit #lim_(x->0)(cos(x)-1)/x#? What is the limit of #sin(2x)/x^2# as x approaches 0? Question #99ee1 What is the derivative of #2^sin(pi*x)#? What is the derivative of #sin^3x#? Question #eefeb Question #af14f See all questions in Limits Involving Trigonometric Functions Impact of this question 1790 views around the world You can reuse this answer Creative Commons License