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

  • Thread starter kingstrick
  • Start date
In summary, by using the theorem that if the limit of a sequence Pn is c and the limit of another sequence Sn is d, then the limit of Pn*Sn is cd, we can prove that if the limit of the sequence Pn goes to x, then the limit of Pn^2 goes to x^2. This is shown by showing that the difference between Pn^2 and x^2 approaches zero as n approaches infinity.
  • #1
kingstrick
108
0

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
 
Physics news on Phys.org
  • #2
kingstrick said:

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|##?
 
  • #3
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
kingstrick said:
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)##.
 
  • #5
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
kingstrick said:
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.
 
  • #7
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
kingstrick said:
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?
 
  • #9
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
kingstrick said:
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?
 
  • #11
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
kingstrick said:
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?
 
  • #13
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
kingstrick said:
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.
 
  • Like
Likes 1 person
  • #15
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
kingstrick said:
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.
 

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

1. What is the definition of a limit?

The limit of a sequence, Pn, as n approaches a value, x, is the value that Pn gets closer and closer to as n gets larger and larger. In other words, it is the value that Pn "approaches" but may never actually reach.

2. How do you prove the limit of Pn goes to x?

To prove the limit of Pn goes to x, we need to show that for any ε > 0 (where ε is a small positive number), there exists an N such that for all n > N, the distance between Pn and x is less than ε. This can be written as |Pn - x| < ε.

3. What is the relation between the limit of Pn goes to x and the limit of Pn^2 goes to x^2?

The limit of Pn^2 goes to x^2 is essentially the same as the limit of Pn goes to x, but the function is squared. This means that the value of Pn^2 will get closer and closer to x^2 as n gets larger, just like how Pn gets closer to x. In other words, if the limit of Pn goes to x, then the limit of Pn^2 goes to x^2 as well.

4. What is the proof for the limit of Pn^2 goes to x^2?

To prove the limit of Pn^2 goes to x^2, we can use the definition of a limit and the properties of limits. First, we can write Pn^2 as Pn * Pn. Then, we can use the properties of limits to show that the limit of Pn * Pn is equal to the limit of Pn multiplied by the limit of Pn, which is x * x = x^2. Therefore, the limit of Pn^2 goes to x^2.

5. Can the limit of Pn^2 goes to x^2 be proven using other methods?

Yes, there are multiple methods for proving the limit of a sequence. One common method is the epsilon-delta proof, where we show that for any ε > 0, there exists a corresponding δ > 0 such that if |Pn - x| < δ, then |Pn^2 - x^2| < ε. Another method is the squeeze theorem, where we use the fact that if an ≤ bn ≤ cn and the limits of an and cn both go to x, then the limit of bn also goes to x.

Similar threads

Bounded non-decreasing sequence is convergent
Replies
5
Views
2K
Prove: Limit Point of H ∪ K if p is Limit Point of H or K
Replies
5
Views
1K
Proving Limit Exists: x-2 of f(x)=2
Replies
2
Views
1K
Real analysis: prove the limit exists
Replies
13
Views
2K
Very tricky problem from Spivak's Calculus, Ch. 4 & 5: Graphs/Limits
Replies
32
Views
2K
Proving discontinuity for rational numbers (reduced form)
Replies
3
Views
2K
Limit and Integration of ##f_n (x)##
Replies
8
Views
1K
Help me prove integral answer over infinitesimal interval
Replies
9
Views
1K
The integral of (sin x + arctan x)/x^2 diverges over (0,∞)
Replies
5
Views
2K
Prove that the limit of |x|/x at x=0 DNE
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top