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# Answer to Question #6507 in Real Analysis for Phu Khat Nwe

Question #6507
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),
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).

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