Convergence Proof: Showing (\sqrt{x_{n}})\rightarrow0

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Homework Statement


Let [tex]x_{n\geq}[/tex]0 for all n in the natural numbers.
If ([tex]x_{n}[/tex])[tex]\rightarrow[/tex]0, show that ([tex]\sqrt{x_{n}}[/tex])[tex]\rightarrow[/tex]0.



Homework Equations





The Attempt at a Solution


So far, I have started with [tex]\left|\sqrt{x_{n}}-0\right|[/tex]. Not sure if that's the right way to start.
 
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You should probably start with the definition of convergence
 
A sequence converges to a real number a if for every positive [tex]\epsilon[/tex], there exists an N element of the natural numbers such that whenever n[tex]\geq[/tex]N, it follows that [tex]\left|a_{n}-a\right|[/tex]<[tex]\epsilon[/tex].
 
[tex]a^{2} \leq b^{2}[/tex] iff [tex]a \leq b[/tex]

[tex]a,b \geq 0[/tex]

Can you use this ?
 
Then the sequence is less than 0 and thus converges to 0
 
How did you arrive at such a conclusion? Btw what you said not correct.

How is the sequence less than zero ? In your definition [tex]x_{n} \geq 0[/tex].
 
so the sequence is greater than 0 because x is greater than 0.
 
All I wanted you to do wanted you to do was take the square root of both sides of the inequality...
[tex]x_{n} < \epsilon[/tex].
 
If g(x) --> A when x --> a then p(g(x)) --> p(A) when x --> a.
 
Hmm... what if p was the square root function and A was negative. ?
 
Well A needs to be in the domain of p for it to make sense. Guess I should have written that..
 
It's fine. Btw this theorem is not one of the 4 limit theorems given in most analysis books so I doubt the OP can use it. OP would need to prove it to use it.
 
We had 5 limit theorems when I did analysis in first year at uni. The proof is like 3 lines and not harder than the rest so I think it is strange it isn't standard at other places
 
I think you are referring to limits of functions not limit theorems. The limit theorems are for sequences and they are later generalized to functions.
I am taking analysis right now and the thoerem you mentioned is in the limit of functions section