Question

How do you find the limit of `(x^2 - 8) / (8x-16)` as x approaches `sqrt8^+`?

Answer 1

`lim x->sqrt(8)^+ = 0`

Explanation:

Remember that a limit value is a `y`-value.

Just plug in `x = sqrt(8)` into the function:

`((sqrt(8))^2-8)/(8sqrt(8)-16) = (8-8)/(8sqrt(4)sqrt(2)-16) = 0/(16sqrt(2)-16) = 0`

From the graph you can see that at `sqrt(8) ~~ 2.8284` the function crosses the `y=0` line `(x-"axis"):`
graph{(x^2-8)/(8x-16) [-10, 10, -5, 5]}