Question

What is `lim_(x->2) (x^5-32)/(x-2)` ?

Answer 1

`lim_(x->2) (x^5-32)/(x-2)=80`

Explanation:

Notice that in general:

`a^5-b^5 = (a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)`

So we find:

`x^5-32 = x^5-2^5`

`color(white)(x^5-32) = (x-2)(x^4+2x^3+2^2x^2+2^3x+2^4)`

`color(white)(x^5-32) = (x-2)(x^4+2x^3+4x^2+8x+16)`

So we can simplify our rational expression to find:

`lim_(x->2) (x^5-32)/(x-2) = lim_(x->2) ((color(red)(cancel(color(black)(x-2))))(x^4+2x^3+4x^2+8x+16))/(color(red)(cancel(color(black)(x-2))))`

`color(white)(lim_(x->2) (x^5-32)/(x-2)) = lim_(x->2) (x^4+2x^3+4x^2+8x+16)`

`color(white)(lim_(x->2) (x^5-32)/(x-2)) = (color(blue)(2))^4+2(color(blue)(2))^3+4(color(blue)(2))^2+8(color(blue)(2))+16`

`color(white)(lim_(x->2) (x^5-32)/(x-2))=80`