How do you evaluate the limit #((2+x)^3-8)/x# as x approaches #0#?

2 Answers
Oct 6, 2016

#lim_(x->0) ((2+x)^3-8)/x = 12#

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 #x != 0#

Hence:

#lim_(x->0) ((2+x)^3-8)/x = lim_(x->0) (12+6x+x^2) = 12#

Oct 6, 2016

#12#

Explanation:

#a^3-b^3 = (a^2+a b+b^2)(a-b)#

so

#((2+x)^3-2^3)/x =( ((2+x)^2+(2+x)2+4)(2+x-2))/x#

so

#lim_(x->0)((2+x)^3-2^3)/x = lim_(x->0) ((2+x)^2+(2+x)2+4)=12#