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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.
Expert's answer
a) Suppose xn converges to 0 but sqrt(x(n)) does not converges to
Therefore we can find a subsequence
and d>0 such that

sqrt( x(n_k) ) > d > 0
for all k.

(*) 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

So assume that x>0, and so there exists N>0 such that

x(n) > x/2 for all n>N.


sqrt(x(n)) >

Denote A = sqrt(x(n)) + sqrt(x).
A =
sqrt(x(n)) + sqrt(x) > x/2 + sqrt(x/2) > 0.

Now fix
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.


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
|sqrt(x(n)) - sqrt(x)| < eps
whenever n>N.
So sqrt(x(n))
converges to sqrt(x).

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