Question

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

Answer 1

As `y=sqrtx` is a strictly increasing function:

`0 < x < epsilon^2 <=> 0 < sqrt(x) < sqrt(epsilon^2)`

So, for any `epsilon > 0`, if we take `delta_epsilon = epsilon^2` we have:

`0 < x < delta_epsilon <=> 0 < sqrt(x) < epsilon`

In particular for `epsilon =0.01` we can choose `delta_epsilon <= 0.0001`