How do you find the limit of # (x-pi)/(sinx)# as x approaches pi? Calculus Limits Determining Limits Algebraically 1 Answer James · Jacobi J. May 5, 2018 The answer #lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-1# Explanation: show the steps #lim_(xrarrpi)(x-pi)/(sinx)# Direct compensation product equal #(0/0)# we must use L'Hopital's Rule #lim_(xrarra)[f'(x)]/[g'(x)]# if the direct compensation product equal #(0/0)# in your question #f(x)=x-pi# #f'(x)=1# #g(x)=sinx# #g'(x)=cosx# #lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-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 8708 views around the world You can reuse this answer Creative Commons License