How do you evaluate the limit #((2+x)^3-8)/x# as x approaches #0#?
2 Answers
Oct 6, 2016
Explanation:
#((2+x)^3-8)/x = (2^3+3(2^2)x+3(2)x^2+x^3-8)/x#
#color(white)(((2+x)^3-8)/x) = (color(red)(cancel(color(black)(8)))+12x+6x^2+x^3-color(red)(cancel(color(black)(8))))/x#
#color(white)(((2+x)^3-8)/x) = ((12+6x+x^2)color(red)(cancel(color(black)(x))))/color(red)(cancel(color(black)(x)))#
#color(white)(((2+x)^3-8)/x) = 12+6x+x^2#
with exclusion
Hence:
#lim_(x->0) ((2+x)^3-8)/x = lim_(x->0) (12+6x+x^2) = 12#
Oct 6, 2016
Explanation:
so
so