Answer to Question #6507 in Real Analysis for Phu Khat Nwe
Let xn is greater than or equal to 0 for all n belongs to N.
(a) If xn converges to 0, show that sqrt (xn) converges to 0.
(b) If xn converges to x, show that sqrt (xn) converges to xn.
a) Suppose xn converges to 0 but sqrt(x(n)) does not converges to 0. Therefore we can find a subsequence x(n_k) and d>0 such that
sqrt( x(n_k) ) > d > 0 for all k.
Then (*) x(n_k) > d^2 > 0
On the other hand, since (x(n)) converges to 0, we can find N>0 such that d^2 > x(n) for all n>N. In particular, if n_k > N, then (**) d^2 > x(n_k), which contradicts to (*), and so to the existence of the subsequence (x(n_k)).
Hence sqrt(x(n)) converges to 0.
b) Suppose x(n) converges to x. We have to show that sqrt(x(n)) converges to sqrt(x). If x=0, then teh result is proved in a).
So assume that x>0, and so there exists N>0 such that
x(n) > x/2 for all n>N.
sqrt(x(n)) > sqrt(x/2)
Denote A = sqrt(x(n)) + sqrt(x). Then A = sqrt(x(n)) + sqrt(x) > x/2 + sqrt(x/2) > 0.
Now fix eps>0. Since x(n) converges to x we can increase N (if necessary) and also assume that |x(n)-x| < A * eps for all n>N.