# Homework Help: Pn->x pn^2 -> x^2

1. Oct 25, 2013

### kingstrick

1. The problem statement, all variables and given/known data

If the limit of the sequence Pn goes to x, prove that the limit of Pn^2 goes to x^2

2. Relevant equations

can't use epsilon and delta

3. The attempt at a solution

I thought simply showing that arbitrary intervals work is enough. but my teacher says that my solution doesn't make sense. any help as to my misconception would make sense.

Let O be any open interval defined as (√a,√b) such that x is in the interval and x is the limit point of Pn. then √a < x < √b. then a < x^2 < b. Then for any open interval O1, x^2 is contained. Thus lim Pn^2 is X^2

2. Oct 25, 2013

### LCKurtz

Why don't you try looking at $|p_n^2-x^2|$ and use what you know about $|p_n-x|$?

3. Oct 25, 2013

### kingstrick

hmmmm....

Can I say that:

let Pn = Sn, where the limit of Pn = x, then lim Sn = x. Then lim (Pn*Sn) = x^2. Then lim Pn^2 = x^2.

This makes sense to me, but feels like i am cheating.

4. Oct 25, 2013

### Dick

What is Sn?? You are cheating. LCKurtz is just suggesting you use $p_n^2-x^2=(p_n-x)(p_n+x)$.

5. Oct 25, 2013

### kingstrick

I was trying to define another sequence. I apologize for my poor form. This stuff has me totally overwhelmed.

I thought if i defined two equivalent sequences then they would have the same limit. Then i could use like the multiplication of limits to show that it is x^2

6. Oct 25, 2013

### Dick

That's ok. I actually stand corrected. If you have the theorem that $lim(P_n)=c$ and $lim(S_n)=d$ then $lim(P_nS_n)=cd$ then that works fine. It would only be cheating if you don't have that theorem yet. Sorry.

7. Oct 26, 2013

### kingstrick

oh. I didn't know we had to prove this. It just seemed logical. I guess i will just have to go back to the drawing board. I see where you were going for in the previous post given that it factors and i can show they both have |Pn-x| but i don't know what to do with the other part since there is an addition.

8. Oct 26, 2013

### Dick

Yes, it is logical but you do have to prove it before you can take that route. But in the meantime |Pn-x| converges to zero. What does |Pn+x| converge to?

9. Oct 26, 2013

### kingstrick

Here is my shameful ignorance again. I thought that |Pn-L| meant that Pn converges to L, so |Pn-x| converges to x and |Pn+x| converges to -x. I don't see how you get zero. but that would make |Pn-x||Pn+x| converging to
-x*x which is -x^2 ... so i know i am messing up somewhere.

thank you for your patience btw.

10. Oct 26, 2013

### Dick

I've got all the patience in the in the world, to a certain degree. Pn converging to x means for very large n, Pn is almost equal to x. Wouldn't that mean |Pn-x| is almost zero?

11. Oct 26, 2013

### kingstrick

Yes. I see that now. So |Pn-x| = 0 then |Pn+x| = 2x? but since (Pn-x)(Pn+x) wouldn't that also go to zero? since i would essentially be having zero times 2x for very large Pn.

12. Oct 26, 2013

### Dick

|Pn-x| isn't equal to zero, but it approaches zero. And yes, |Pn+x| approaches |2x| which is a finite number. So |Pn^2-x^2| must also approach 0. So lim(Pn^2) must be what?

13. Oct 26, 2013

### kingstrick

So...
since lim Pn goes to x as n goes to infinity,
then
|pn-x| goes to zero as n goes to infinity,
then
(Pn-x)(Pn+x) goes to (0)(2x) goes to 0 as n goes to infinity
then
(Pn^2 - x^2)= (Pn-x)(Pn+x) goes to 0 as n goes to infinity
then |Pn^2-X^2| goes to 0 as n goes to infinity
therefore the limit of Pn^2 = x^2

did i finally get it?

14. Oct 26, 2013

### Dick

Sure you did. And I think that's probably exactly what they wanted you to do.

15. Oct 26, 2013

### kingstrick

now once i prove this in class, does that "open the door" for the use of the multiplication trick? or is that a separate rule?

16. Oct 26, 2013

### Dick

Now that is a REALLY intelligent question! I think you are getting this whole limit thing. I was just asking myself the same thing. This is a special case of the multiplication trick. It's easier to prove and maybe a little more obvious but it still needs to be proved. So it's not a separate rule. But I'd wait for it to be proved in class before you "open the door", for pedagogical reasons. Then your proof in post 3 is perfectly valid.