Real Analysis (Sequences)

**Phyxius117** · September 19th 2010, 11:38 AM

Here is the problem. How would I prove this problem?

**Plato** · September 19th 2010, 11:48 AM

Is it clear that $\sqrt{x_n}-\sqrt{a}=\dfrac{x_n-a}{\sqrt{x_n}+\sqrt{a}}?$

Can you finish?

**Phyxius117** · September 19th 2010, 02:47 PM

Do I go with a direct proof?

like:

Given $\epsilon$ > 0

Then $|\sqrt(x_n)-\sqrt(a)| < \epsilon$

Thus $|{\frac{x_n-a}{\sqrt(x_n)+\sqrt(a)}| < \epsilon$

but what do I do next? Do I split it up into two?

**Plato** · September 19th 2010, 03:17 PM

Originally Posted by Phyxius117

Given $\epsilon$ > 0
Then $|\sqrt(x_n)-\sqrt(a)| < \epsilon$
Thus $|{\frac{x_n-a}{\sqrt(x_n)+\sqrt(a)}| < \epsilon$

That is pure wishful thinking.
First show that $\left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow x_n > \frac{a}{2}} \right]$

From that you show that
$\frac{1}{{\sqrt {x_n } + \sqrt a }} < \frac{{\sqrt 2 }}<br /> {{\sqrt a + \sqrt {2a} }}$

You have a lot of work to do.

**Phyxius117** · September 19th 2010, 03:24 PM

I don't quite understand. how did you get x_n > a/2 ?

**Plato** · September 19th 2010, 03:28 PM

Originally Posted by Phyxius117

I don't quite understand. how did you get x_n > a/2 ?

I did not prove that!
I said that you must prove that as part of this problem.

Look we are not going to do this for you.
It is your problem. You do it.
I told you how it is done.

**Phyxius117** · September 19th 2010, 04:59 PM

Thank you for all the help.

after looking at your hints on how to approach this problem, i'm trying hard to figure out how to solve it but I just don't understand what I am suppose to do.

**magus** · September 20th 2010, 01:37 AM

I happened on this problem and I was wondering if the following reasoning is enough to prove that $x_n>\frac{a}{2}$ .

If $x_n$ is monotonically increasing or non-decreasing then it must eventually pass $\frac{a}{2}$ in order to get to $a$ . If it is monotonically decreasing or non-incresing then all values of $x_n$ must be greater then $a$ . If it is none of those we can still choose a monotonic subsequence because all sequences have a monotonic subsequence and apply the logic previously stated.

Is this enough just for $x_n>\frac{a}{2}$ ?

**MathoMan** · September 20th 2010, 02:37 AM

Originally Posted by Plato

That is pure wishful thinking.
First show that $\left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow x_n > \frac{a}{2}} \right]$

From that you show that
$\frac{1}{{\sqrt {x_n } + \sqrt a }} < \frac{{\sqrt 2 }}<br /> {{\sqrt a + \sqrt {2a} }}$

You have a lot of work to do.

Hi. I like these assignments, they can be challenging. I am curios Plato if your first hint can be proved like this:

$\lim x_n =a \Rightarrow \forall \epsilon >0, \exists n_\epsilon$ such that $\forall n>n_\epsilon$ is $|x_n-a|<\epsilon.$ Setting $\epsilon =\frac{a}{2}$ then set $N=n_\epsilon=n_{\frac{a}{2}}$ and for all $n>N$ we'd have $|x_n-a|<\epsilon=\frac{a}{2}.$

Now we get
$|x_n-a|<\frac{a}{2},$
$-\frac{a}{2}<x_n-a<\frac{a}{2},$ (add a to all)
$\frac{a}{2}<x_n<\frac{3a}{2},$ and the left part of this extended inequality gives us $x_n>\frac{a}{2}.$

I guess this is ok, Plato can you respond. And am trying to get the second part. Still, I don't see were this leads. Thx in advance.

**MathoMan** · September 20th 2010, 02:59 AM

Continuing on my previous post. If my getting to $x_n>\frac{a}{2}$ is ok then:

$\sqrt{x_n}>\sqrt{\frac{a}{2}},$ (now add $\sqrt{a}$ )
$\sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a},$ (compare the reciprocals)
$\frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\ sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr t{2a}}$

So we get the $\frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}}{\sqr t{a}+\sqrt{2a}}.$ Now what?

EDIT: see my next post, I almost got it

**MathoMan** · September 20th 2010, 03:17 AM

I thought I should put it all in one place.

$\lim x_n =a \Rightarrow \forall \epsilon >0, \exists n_\epsilon$ such that $\forall n>n_\epsilon$ is $|x_n-a|<\epsilon.$ Setting $\epsilon =\frac{a}{2}$ then set $N=n_\epsilon=n_{\frac{a}{2}}$ and for all $n>N$ we'd have $|x_n-a|<\epsilon=\frac{a}{2}.$

Now we get
$|x_n-a|<\frac{a}{2},$
$-\frac{a}{2}<x_n-a<\frac{a}{2},$ (add a to all)
$\frac{a}{2}<x_n<\frac{3a}{2},$ and the left part of this extended inequality gives us $x_n>\frac{a}{2}.$
Then,
$\sqrt{x_n}>\sqrt{\frac{a}{2}},$ (now add $\sqrt{a}$ )
$\sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a},$ (compare the reciprocals)
$\frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\ sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr t{2a}}$

So we get the $\frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}}{\sqr t{a}+\sqrt{2a}}.$

If we were to multiply that with $x_n-a$
finally I see why $\sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}$ comes in handy.

$\sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}$

Forcing limit on the above should lead towards solution. Well almost. Strict inequality is bothering me here. Anyway Physix117 that started the thread should work out the fine details. I gave my shot at the rough stuff and I think I did just fine. Waiting for feedback.

**HallsofIvy** · September 20th 2010, 04:46 AM

Physix117, one problem with people not showing any work in their first post is that there are often many different ways to solve a problem, of differing difficulty and mathematical sophistication- and we have no way of knowing which is appropriate for you.

Plato chose to show you the most basic- and therefore, most difficult, method, assuming the least he could about your knowledge- that if you are asked to find limits, you at least know the definition of "limit". If I were doing a problem like this, I think I would just observe that $\sqrt{x}$ , for x> 0, is a continuous function and so $\limit_{n\to\infty} \sqrt{x_n}= \sqrt{\lim_{n\to\infty} x_n}= \sqrt{a}$ .

Mathoman, "strict inequality" should not bother you. If $a_n< B$ for all n, it is NOT true that $\lim_{n\to\infty} a_n< B$ . What is true is that $\lim_{n\to\infty}\le B$ .

Also, $x_n> \frac{a}{2}$ is not always true nor is it claimed to be. For "sufficiently large n", $x_n> \frac{a}{2}$ . That is, "there exist a positive integer N such that if n> N, then $x_n> \frac{a}{2}$ ". That follows directly from the definition of "limit", taking $\epsilon= \frac{a}{2}> 0$ . Since $\lim_{n\to\infty} x_n= a$ , given any $\epsilon> 0$ there exist N such that if n> N, then $|x_n- a|< \epsilon$ . Taking $\epsilon= \frac{a}{2}$ that becomes $|x_n- a|< \frac{a}{2}$ which is the same as $-\frac{a}{2}< x_n- a< \frac{a}{2}$ . Adding a to each part, $a- \frac{a}{2}= \frac{a}{2}< x_n< a+ \frac{a}{2}= \frac{3a}{2}$ . Of course, it is only the left part you need here.

**magus** · September 21st 2010, 04:24 PM

Halls of Ivy and Mathoman. I can see now where plato was leading but there's something that bothers me about the rhs. It involves $x_n$ and I can't get it in a form leading to $|\sqrt{x_n}-\sqrt{a}|$ so I was wondering if something different would work.

Starting from
$\sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}$

$lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq lim\limits_{n\to\infty} \frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}$

$lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq \frac{\sqrt{2}lim\limits_{n\to\infty}(x_n-a)}{\sqrt{a}+\sqrt{2a}}$

$lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq \frac{0}{\sqrt{a}+\sqrt{2a}}$
$lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq 0$

Is this valid and does it help any? I figure that from the original frac{a}{2}<x_n<\frac{3a}{2} I could find a lower bound with a $(x_n-a)$ term in the numerator that would make the limit of it equal 0 and then since $0 \leq lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq 0$

We'd know that $lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} =0$

Is this at all valid?

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