Question

How do you find the limit of `(sqrtx-2)/(x-4)` as `x->4^-`?

Answer 1

`lim_(x->4^-)(sqrtx-2)/(x-4)=1/4`

Explanation:

Since the function,
as `x->4^-`, is equal to the indeterminate form: `0/0`

We can apply L'Hopital's Rule:

Therefore:

`lim_(x->4^-)(sqrtx-2)/(x-4)=>0/0`

`=^(L'H) lim_(x->4^-)(1/2x^(-1/2))=1/(2sqrt4)=1/4`