What is #lim_(x->2) (x^5-32)/(x-2)# ?

1 Answer
Oct 8, 2016

Answer:

#lim_(x->2) (x^5-32)/(x-2)=80#

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#