Limit of Pn Goes to x, Prove Limit of Pn^2 Goes to x^2

[kingstrick]

Homework Statement

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

Homework Equations

can't use epsilon and delta

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

can't use epsilon and delta

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

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

 

[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.

 

[Dick]

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.

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

 

[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

 

[Dick]

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

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.

 

[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.

 

[Dick]

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.

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?

 

[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.

 

[Dick]

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.

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?

 

[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.

 

[Dick]

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.

|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?

 

[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?

 

[Dick]

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?

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

 

[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?

 

[Dick]

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?

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.