Question

What is the limit of `[x^2-4x+4]/[x^3-64]` as x goes to infinity?

Answer 1

It is `0`.

Explanation:

`lim_(xrarroo) [x^2-4x+4]/[x^3-64]` has indeterminate form `oo/oo`

Make the denominator not go to `oo`:

`lim_(xrarroo) [x^2-4x+4]/[x^3-64] = lim_(xrarroo) [x^3(1/x-4/x^2+4/x^3)]/[x^3(1-64/x^3)`

`= lim_(xrarroo) (1/x-4/x^2+4/x^3)/(1-64/x^3)`

`= (0-0+0)/(1-0) = 0`