Question 3.9: If xn ≥ 0 for all n ∈ N and lim xn = x, then lim √xn = √x.

Let x = 0. If ε is a positive number then, working with ε^{2}, we can find an N ∈ \mathbb{N} such that

|x_{n} − 0| =x_{n} < ε^{2}  for all n ≥ N.

But this implies \left|\sqrt{x_{n}}− 0\right| = \sqrt{x_{n}} < ε for all n ≥ N. Hence \sqrt{x_{n}} → 0.

If x ≠ 0 then x > 0 and hence \sqrt{x} > 0. In this case

0 ≤ \left|\sqrt{x_{n}} − \sqrt{x} \right| = \frac{\left|x_{n} − x\right|}{\left|\sqrt{x_{n}} + \sqrt{x}\right|}≤ \frac{\left|x_{n} − x\right|}{\sqrt{x}},

and, since lim|x_{n} − x| = 0, Theorem 3.6 implies lim \left|\sqrt{x_{n}} − \sqrt{x}\right| = 0, or lim \sqrt{x_{n}} = \sqrt{x}.