Question

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

Answer 1

`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`

Answer 2

`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`