Question

What is `lim x->2` of `(x^3-8)/(x-2)`?

Answer 1

`12`

Explanation:

`lim_(x->2) (x^3-8)/(x-2)`

it's `0/0` indeterminate so you can l'Hopital it, if you like, but we can pursue it in other ways.

if we set `h = x - 2` or `x = 2+h`, where clearly `0< abs(h) "<<" 1`

then `lim_(x->2) (x^3-8)/(x-2)`

`= lim_(h->0) ((2+h)^3-8)/(h)`

`= lim_(h->0) ((2(1+h/2))^3-8)/(h)`

`= lim_(h->0) (8(1+h/2)^3-8)/(h)`

and by binomial expansion
`= lim_(h->0) (8(1+3* h/2 + O(h^2))-8)/(h)`

`= lim_(h->0) (8+12 h + O(h^2)-8)/(h)`

`= lim_(h->0) 12 + O(h)`

`=12`

Answer 2

`12`

Explanation:

Recall the difference of cubes identity:

`a^3-b^3=(a-b)(a^2+ab+b^2)`

Note, too, that the numerator is expressible as a difference of cubes:

`x^3-8=x^3-2^3=(x-2)(x^2+x(2)+2^2)`

`=(x-2)(x^2+2x+4)`

So, we see that:

`lim_(xrarr2)(x^3-8)/(x-2)=lim_(xrarr2)(color(red)cancel(color(black)((x-2)))(x^2+2x+4))/color(red)cancel(color(black)((x-2)`

`=lim_(xrarr2)(x^2+2x+4)`

`=2^2+2(2)+4`

`=4+4+4`

`=12`