Question

How do you use the epsilon delta definition to prove that the limit of `sqrt(x+6)=9` as `x->3`?

Answer 1

For each epsilon, we can always choose `delta < (9 + epsilon)^2 - 9`

Explanation:

`forall epsilon >0, exists delta >0 ; | x - 3| < delta => |sqrt{x + 6} - 9| < epsilon`

I will think backwards

`-epsilon < sqrt{x + 6} - 9 < epsilon`

plus nine

`9-epsilon < sqrt{x + 6} < 9 + epsilon`

^2

`(9-epsilon)^2 < x + 6 < (9 + epsilon)^2`

minus 9

`(9-epsilon)^2 - 9 < x - 3 < (9 + epsilon)^2 - 9`

Why? All we want is `-delta < x - 3 < delta`

We can always choose `delta < (9 + epsilon)^2 - 9`

And `(9 - epsilon)^2 - 9 < - delta`

`9 - (9 - epsilon)^2 < delta < (9 + epsilon)^2 - 9`

`9 - (9 - epsilon)^2 < (9 + epsilon)^2 - 9`

#18 < (9 + epsilon)^2 + (9 - epsilon)^2 = 81 + 81 +(18-18)epsilon + 2epsilon^2#

`9-81 = -72 < epsilon^2`