How do you evaluate the limit of #lim (2x-8)/(x^3-64)# as #x->4#? Precalculus Limits Two-Sided Limits 1 Answer Shwetank Mauria Jan 18, 2017 #Lt_(x->4)(2x-8)/(x^3-64)=1/24# Explanation: #Lt_(x->4)(2x-8)/(x^3-64)# and as #x^3-64=x^3-4^3#, using identity #(a^3-b^3)=(a-b)(a^2+ab+b^2)#, the above can be written as = #Lt_(x->4)(2(x-4))/((x-4)(x^2+4x+16))# = #Lt_(x->4)2/(x^2+4x+16)# = #2/(4^2+4xx4+16)# = #2/(16+16+16)# = #1/24# 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 3059 views around the world You can reuse this answer Creative Commons License