Question

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.

Answer 1

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.

Then

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.


Notice
that

x(n) - x = ( sqrt(x(n)) + sqrt(x) ) * ( sqrt(x(n)) - sqrt(x)
),

whence for n>N we have that
x(n) - x > A * ( sqrt(x(n)) -
sqrt(x) )

and so
|sqrt(x(n)) - sqrt(x)| < |x(n) - x| / A < A
* eps /A = eps.


Thus for any eps>0 we can find N>0 such
that
|sqrt(x(n)) - sqrt(x)| < eps
whenever n>N.
So sqrt(x(n))
converges to sqrt(x).