3.2.10 Theorem Let X = (xn) be a sequence of real numbers that converges to x and suppose that xn ≥ 0. Then the sequence (√xn) of positive square roots converges and lim(√xn) = √x.
Proof. It follows from Theorem 3.2.4 that x = lim(xn) ≥ 0 so the assertion makes sense. We now consider the two cases: (i) x = 0 and (ii) x > 0.
Case (i) If x = 0, let ε > 0 be given. Since xn → 0 there exists a natural number K such that if n ≥ K then
0 ≤ xn = xn - 0 < ε².
Therefore [see Example 2.1.13(a)], 0 ≤ √xn < ε for n ≥ K. Since ε > 0 is arbitrary, this implies that √xn → 0.
Case (ii) If x > 0, then √x > 0 and we note that
√xn - √x = (√xn - √x)(√xn + √x) / (√xn + √x) = (xn - x) / (√xn + √x)
Since √xn + √x ≥ √x > 0, it follows that
|√xn - √x| ≤ (1 / √x) |xn - x|.
The convergence of √xn → √x follows from the fact that xn → x. Q.E.D.