How do you find the limit of #cosx/cotx# as #x->pi/2#? Calculus Limits Determining Limits Algebraically 1 Answer Mia Nov 14, 2016 #lim _(x->pi/2) cos x / cot x = 1# Explanation: #lim _(x->pi/2) cos x / cot x# #" "# #= lim_(x->pi/2) cosx/(cosx/sinx)# #" "# #" "# #= lim_(x->pi/2)cosx/1 xx sinx/cosx# #" "# #" "# #= lim_(x->pi/2)cancel(cosx)/1 xx sinx/cancel(cosx)# #" "# #=lim_(x->pi/2)sinx# #" "# #=sin (pi/2)# #" "# #=1# Answer link Related questions How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ? How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ? How do you find the limit #lim_(x->4)(x^3-64)/(x^2-8x+16)# ? How do you find the limit #lim_(x->2)(x^2+x-6)/(x-2)# ? How do you find the limit #lim_(x->-4)(x^2+5x+4)/(x^2+3x-4)# ? How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ? How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ? How do you find the limit #lim_(h->0)((2+h)^3-8)/h# ? How do you find the limit #lim_(x->9)(9-x)/(3-sqrt(x))# ? How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ? See all questions in Determining Limits Algebraically Impact of this question 12149 views around the world You can reuse this answer Creative Commons License