# square root function and convergence

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• Jun 1st 2009, 03:19 PM
pberardi
square root function and convergence
This one is giving me such a hard time. I guess this is important to know because it is saying that the square root of the function translates to the square root of the limit. Similar to mutliplication and addition. The hint they give is to use the fact that if all convergent subsequences have the same limit, then the sequence is convergent. I need help on this one. Thanks.

Suppose <x_n> --->x for each natural number n, where x_n >= 0. Prove that <sqrt(x_n)> ---> to sqrt(x)
• Jun 1st 2009, 07:57 PM
nqramjets
Perhaps this is not what you're looking for, but sqrt is continuous, so the limit of f(x_n)=f(lim x_n). Or are you confused about epsilon delta?
• Jun 1st 2009, 08:00 PM
pberardi
Quote:

Originally Posted by nqramjets
Perhaps this is not what you're looking for, but sqrt is continuous, so the limit of f(x_n)=f(lim x_n). Or are you confused about epsilon delta?

I have to prove this for all x_n>= 0. I probably have to use a convergence proof or use the squeeze theorem. Any suggestions?
• Jun 1st 2009, 08:04 PM
nqramjets
I think I'm confused by what you mean... Do you mean that you have a sequence $(x_n)\rightarrow x$ and you're trying to show that $(\sqrt{x_n})\rightarrow \sqrt{x}$ or something else?
• Jun 1st 2009, 08:06 PM
pberardi
Quote:

Originally Posted by nqramjets
I think I'm confused by what you mean... Do you mean that you have a sequence $(x_n)\rightarrow x$ and you're trying to show that $(\sqrt{x_n})\rightarrow \sqrt{x}$ or something else?

Exactly that. Nothing else.
• Jun 1st 2009, 08:13 PM
nqramjets
First show that the square root is continuous:
Let $\varepsilon >0$, then given some $x\in(0,\infty)$ choose $\delta < \varepsilon$. Now, consider $y\in(0,\infty)$ such that $|x-y|<\delta$, then
$|\sqrt{x}-\sqrt{y}|\leq |\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|=|x-y|<\delta<\varepsilon$, So square root is continuous.
Now, by continuity, if $x_n\rightarrow x$ for $x_n\in(0,\infty)$ with $x\in(0,\infty)$ you have
$\lim_{n\rightarrow \infty}\sqrt{x_n}=\sqrt{\lim_{n\rightarrow \infty}x_n}=\sqrt{x}$
as desired. Hope this helps!
• Jun 1st 2009, 08:17 PM
pberardi
Helps alot. Thanks bud.
• Jun 1st 2009, 08:36 PM
Random Variable
The following proof is valid if x>0. You can prove the case when x=0 separately.

Assume ( $x_{n}$) converges to x and that $x_{n} \ge 0$

$| \sqrt{x_n} - \sqrt{x} | = \frac { |\sqrt{x_n} - \sqrt{x}||\sqrt{x_n} + \sqrt{x}|}{\sqrt{x_n} + \sqrt{x}}$

$= \frac {|x_{n}-x|}{ \sqrt{x_{n}} + \sqrt{x}}$

$\le \frac {1}{\sqrt{x}}|x_{n}-x|$

And since lim $(x_{n}-x) = 0$, and since $\frac {1}{\sqrt{x}} > 0$, you can conclude that lim $\sqrt{x_{n}} = \sqrt{x}$
• Jun 1st 2009, 08:38 PM
pberardi
This is awesome RandomVariable. Pardon me but how would one begin the case for zero? And also, what does this proof have anything to do with the fact that x_n converges to x?
• Jun 1st 2009, 08:57 PM
Random Variable
Quote:

And also, what does this proof have anything to do with the fact that x_n converges to x?
It's used at the end.

I need to show that $| \sqrt{x_n} - \sqrt{x} | \le k|a_{n}|$ for some k>0 and for some series $a_{n}$ that converges to zero.
• Jun 1st 2009, 09:05 PM
Random Variable
I think you should be able to prove the the case for when x=0 straight from the definition. And I emphasize "think" because I can't seem to prove it. (Worried)
• Jun 2nd 2009, 08:43 AM
nqramjets
I'm pretty sure the proof I posted above actually works just fine for x=0. That affects nothing in the proof. I mean, sqrt is certainly continuous at x=0, so you are good.
• Jun 2nd 2009, 12:10 PM
Random Variable
I think I came up with a proof for the case when x=0.

Assume $x_{n} \ge 0$

$(x_{n}) \to 0 \Longrightarrow \forall \epsilon >0, \ \exists N : \forall n>N, \ |x_{n}-0| = |x_{n}| = x_{n} < \epsilon^{2}$

then $|\sqrt{x_{n}} -0| = |\sqrt{x_{n}}| = \sqrt{x_{n}} < \sqrt{ \epsilon^{2}} = \epsilon$
• Jun 2nd 2009, 01:05 PM
nqramjets
You can use the continuous inverse theorem on $f(x)=x^2$ on $[0,\infty)$ to get continuity for the whole thing. This result follows from the preservation of interval theorem.

By the way, I think you need to follow the logic on the second proof provided here, because mine actually has a fault in it. I didn't even think about the fact that $\sqrt{x}+\sqrt{y} < 1$ in which case that inequality isn't true... So yea, go with the mult/div by that expression to fix it. Sorry!
• Jun 2nd 2009, 01:21 PM
Random Variable
So my proof for the case when x=0 is valid?
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