# Thread: square root function and convergence

1. ## 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)

2. 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?

3. 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?

4. 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?

5. 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.

6. This isn't so bad:
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!

7. Helps alot. Thanks bud.

8. 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 $\sqrt{x_{n}} = \sqrt{x}$

9. 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?

10. 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.

11. 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.

12. 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.

13. 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 $|\sqrt{x_{n}} -0| = |\sqrt{x_{n}}| = \sqrt{x_{n}} < \sqrt{ \epsilon^{2}} = \epsilon$

14. 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!

15. So my proof for the case when x=0 is valid?

Page 1 of 2 12 Last

### If (x_n) converges to x, show (sqrt x_ n) converges to x

Click on a term to search for related topics.