Question

What is `lim x->64` of `(sqrtx -8)/(x-64)`?

Answer 1

The answer is `=1/16`

Explanation:

`lim_(x->64)(sqrtx-8)/(x-64)=(8-8)/(64-64)=0/0`

This is undefined

So,

we apply L'Hôpital's rule

`lim_(x->64)(sqrtx-8)/(x-64)=lim_(x->64)((sqrtx-8)')/((x-64)')`

`=lim_(x->64)(1/(2sqrtx))/(1)=1/(2sqrt64)=1/16`