How do you evaluate the limit #((x+1)^3-1)/x# as x approaches #0#?
1 Answer
Nov 8, 2016
Explanation:
#lim_(x->0) ((x^3+1)^3-1)/x = lim_(x->0) ((x^9+3x^6+3x^3+color(red)(cancel(color(black)(1))))-color(red)(cancel(color(black)(1))))/x#
#color(white)(lim_(x->0) ((x^3+1)^3-1)/x) = lim_(x->0) (x^9+3x^6+3x^3)/x#
#color(white)(lim_(x->0) ((x^3+1)^3-1)/x) = lim_(x->0) (x^8+3x^5+3x^2)#
#color(white)(lim_(x->0) ((x^3+1)^3-1)/x) = 0#