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

http://i10.photobucket.com/albums/a1.../problem46.jpg

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- Sep 19th 2010, 11:38 AMPhyxius117Real Analysis (Sequences)
Here is the problem. How would I prove this problem?

http://i10.photobucket.com/albums/a1.../problem46.jpg - Sep 19th 2010, 11:48 AMPlato
Is it clear that $\displaystyle \sqrt{x_n}-\sqrt{a}=\dfrac{x_n-a}{\sqrt{x_n}+\sqrt{a}}?$

Can you finish? - Sep 19th 2010, 02:47 PMPhyxius117
Do I go with a direct proof?

like:

Given $\displaystyle \epsilon$ > 0

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

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

but what do I do next? Do I split it up into two? - Sep 19th 2010, 03:17 PMPlato
That is pure wishful thinking.

First show that $\displaystyle \left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow x_n > \frac{a}{2}} \right]$

From that you show that

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

{{\sqrt a + \sqrt {2a} }}$

You have a lot of work to do. - Sep 19th 2010, 03:24 PMPhyxius117
I don't quite understand. how did you get x_n > a/2 ?

- Sep 19th 2010, 03:28 PMPlato
- Sep 19th 2010, 04:59 PMPhyxius117
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. - Sep 20th 2010, 01:37 AMmagus
I happened on this problem and I was wondering if the following reasoning is enough to prove that $\displaystyle x_n>\frac{a}{2}$.

If $\displaystyle x_n$ is monotonically increasing or non-decreasing then it must eventually pass $\displaystyle \frac{a}{2}$ in order to get to $\displaystyle a$. If it is monotonically decreasing or non-incresing then all values of $\displaystyle x_n$ must be greater then $\displaystyle a$. 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$\displaystyle x_n>\frac{a}{2}$? - Sep 20th 2010, 02:37 AMMathoMan
Hi. I like these assignments, they can be challenging. I am curios Plato if your first hint can be proved like this:

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

Now we get

$\displaystyle |x_n-a|<\frac{a}{2},$

$\displaystyle -\frac{a}{2}<x_n-a<\frac{a}{2},$ (add*a*to all)

$\displaystyle \frac{a}{2}<x_n<\frac{3a}{2},$ and the left part of this extended inequality gives us $\displaystyle x_n>\frac{a}{2}.$

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. - Sep 20th 2010, 02:59 AMMathoMan
Continuing on my previous post. If my getting to $\displaystyle x_n>\frac{a}{2}$ is ok then:

$\displaystyle \sqrt{x_n}>\sqrt{\frac{a}{2}}, $ (now add $\displaystyle \sqrt{a}$)

$\displaystyle \sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a}, $ (compare the reciprocals)

$\displaystyle \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\ sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr t{2a}}$

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

**EDIT:**see my next post, I almost got it ;) - Sep 20th 2010, 03:17 AMMathoMan
I thought I should put it all in one place.

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

Now we get

$\displaystyle |x_n-a|<\frac{a}{2},$

$\displaystyle -\frac{a}{2}<x_n-a<\frac{a}{2},$ (add*a*to all)

$\displaystyle \frac{a}{2}<x_n<\frac{3a}{2},$ and the left part of this extended inequality gives us $\displaystyle x_n>\frac{a}{2}.$

Then,

$\displaystyle \sqrt{x_n}>\sqrt{\frac{a}{2}}, $ (now add $\displaystyle \sqrt{a}$)

$\displaystyle \sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a}, $ (compare the reciprocals)

$\displaystyle \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\ sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr t{2a}}$

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

If we were to multiply that with $\displaystyle x_n-a$

finally I see why $\displaystyle \sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}$ comes in handy.

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

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. - Sep 20th 2010, 04:46 AMHallsofIvy
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 $\displaystyle \sqrt{x}$, for x> 0, is a**continuous**function and so $\displaystyle \limit_{n\to\infty} \sqrt{x_n}= \sqrt{\lim_{n\to\infty} x_n}= \sqrt{a}$.

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

Also, $\displaystyle x_n> \frac{a}{2}$ is not**always**true nor is it claimed to be. For "sufficiently large n", $\displaystyle x_n> \frac{a}{2}$. That is, "there exist a positive integer N such that if n> N, then $\displaystyle x_n> \frac{a}{2}$". That follows directly from the definition of "limit", taking $\displaystyle \epsilon= \frac{a}{2}> 0$. Since $\displaystyle \lim_{n\to\infty} x_n= a$, given any $\displaystyle \epsilon> 0$ there exist N such that if n> N, then $\displaystyle |x_n- a|< \epsilon$. Taking $\displaystyle \epsilon= \frac{a}{2}$ that becomes $\displaystyle |x_n- a|< \frac{a}{2}$ which is the same as $\displaystyle -\frac{a}{2}< x_n- a< \frac{a}{2}$. Adding a to each part, $\displaystyle a- \frac{a}{2}= \frac{a}{2}< x_n< a+ \frac{a}{2}= \frac{3a}{2}$. Of course, it is only the left part you need here. - Sep 21st 2010, 04:24 PMmagus
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 $\displaystyle x_n$ and I can't get it in a form leading to $\displaystyle |\sqrt{x_n}-\sqrt{a}|$ so I was wondering if something different would work.

Starting from

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

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

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

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

$\displaystyle lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq 0$

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 $\displaystyle (x_n-a)$ term in the numerator that would make the limit of it equal 0 and then since $\displaystyle 0 \leq lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq 0$

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

Is this at all valid?