# Answer on Real Analysis Question 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) 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

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).

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).

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