How do you evaluate the limit of #lim (x^2-25)/(x-5)# as #x->5#? Precalculus Limits Two-Sided Limits 1 Answer bp Nov 29, 2016 10 Explanation: #lim_(x->5) (x^2-25)/(x-5) = lim_(x->5) ((x-5)(x+5))/(x-5)= lim_(x->5) (x+5) = 10# Answer link Related questions What is a two-sided limit? How do I find two-sided limits? What is a limit from below? How do you find limits on a graphing calculator? What are some sample limit problems? What is the limit as #t# approaches 0 of #(tan6t)/(sin2t)#? What is the limit as #x# approaches 0 of #1/x#? What is the limit as #x# approaches 0 of #tanx/x#? Is there a number "a" such that the equation below exists? If so what is the value of "a" and its limit. What is the equation below solved for x to the nearest hundredth? See all questions in Two-Sided Limits Impact of this question 2401 views around the world You can reuse this answer Creative Commons License