How do you evaluate the limit lim(x^2-36)/(x+6)dx as x->1? Calculus Limits Determining Limits Algebraically 1 Answer kumail Dec 27, 2016 lim_(x -> 1) (x^2-36) /( x+6) = -5 Explanation: Simply evaluate the expression at 1. (1^2 - 36)/(1+6) = -35/7 = -5 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 4795 views around the world You can reuse this answer Creative Commons License