What is #lim_(x->2) (x^5-32)/(x-2)# ?
1 Answer
Oct 8, 2016
Explanation:
Notice that in general:
#a^5-b^5 = (a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)#
So we find:
#x^5-32 = x^5-2^5#
#color(white)(x^5-32) = (x-2)(x^4+2x^3+2^2x^2+2^3x+2^4)#
#color(white)(x^5-32) = (x-2)(x^4+2x^3+4x^2+8x+16)#
So we can simplify our rational expression to find:
#lim_(x->2) (x^5-32)/(x-2) = lim_(x->2) ((color(red)(cancel(color(black)(x-2))))(x^4+2x^3+4x^2+8x+16))/(color(red)(cancel(color(black)(x-2))))#
#color(white)(lim_(x->2) (x^5-32)/(x-2)) = lim_(x->2) (x^4+2x^3+4x^2+8x+16)#
#color(white)(lim_(x->2) (x^5-32)/(x-2)) = (color(blue)(2))^4+2(color(blue)(2))^3+4(color(blue)(2))^2+8(color(blue)(2))+16#
#color(white)(lim_(x->2) (x^5-32)/(x-2))=80#
