Question

How do you find the limit of `(x^2 - 8) / (8x-16)` as x approaches `0^-`?

Answer 1

`=1/2`

Explanation:

`lim_(x->0^-) ((x^2-8)/(8x-16))`

`=(0-8)/(0-16)`

`=-8/-16`

`=1/2`