Question

How do you prove that the limit of `sqrtx = 3` as x approaches 9 using the epsilon delta proof?

Answer 1

Start from:

`abs(x-9) = abs(sqrtx-3)abs(sqrtx+3)`

for `x > 0`, so:

`abs(sqrtx-3) = abs(x-9)/(sqrtx+3)`

For any number `epsilon > 0` choose now `delta_epsilon < 3epsilon`.
For `x in (9-delta_epsilon,9+delta_epsilon)` we have;

`abs(x-9) < delta_epsilon`

and, as `sqrtx` is positive:

`sqrtx +3 > 3`

so:

`abs(sqrtx-3) = abs(x-9)/(sqrtx+3) < delta_epsilon/3 < (3epsilon)/3 = epsilon`

Thus:

`x in (9-delta_epsilon,9+delta_epsilon) => abs(sqrtx-3) < epsilon`

which proves that:

`lim_(x->9) sqrtx = 3`