# Math Help - Real Analysis (Sequences)

1. ## Real Analysis (Sequences)

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

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

Can you finish?

3. 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?

4. 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
$\frac{1}{{\sqrt {x_n } + \sqrt a }} < \frac{{\sqrt 2 }}
{{\sqrt a + \sqrt {2a} }}$

You have a lot of work to do.

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

6. 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.

7. 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.

8. 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 $x_n>\frac{a}{2}$?

9. 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
$\frac{1}{{\sqrt {x_n } + \sqrt a }} < \frac{{\sqrt 2 }}
{{\sqrt a + \sqrt {2a} }}$

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} (add a to all)
$\frac{a}{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.

10. 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

11. 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} (add a to all)
$\frac{a}{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.

12. 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.

13. 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?