Question

In the limit `lim sqrt(x^2-4)=0` as `x->2^+`, how do you find `delta>0` such that whenever `2<x<2+delta`, `sqrt(x^2-4)<0.01`?

Answer 1

`delta = 0.00002` would work

Explanation:

Since the convergence to the limit is monotonic, we can solve:

`sqrt(x^2-4) = 0.01`

to find the upper limit for `delta`.

Squaring both sides, this becomes:

`x^2-4 = 0.0001`

Add `4` to both sides to get:

`x^2 = 4.0001`

Then taking the positive square root of both sides, we get:

`x = sqrt(4.0001) ~~ 2.00002499984375195309`

So we could choose `delta = 0.00002`