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

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Homework Statement


Let x_{n\geq}0 for all n in the natural numbers.
If (x_{n})\rightarrow0, show that (\sqrt{x_{n}})\rightarrow0.



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The Attempt at a Solution


So far, I have started with \left|\sqrt{x_{n}}-0\right|. 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 \epsilon, there exists an N element of the natural numbers such that whenever n\geqN, it follows that \left|a_{n}-a\right|<\epsilon.
 
a^{2} \leq b^{2} iff a \leq b

a,b \geq 0

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 x_{n} \geq 0.
 
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...
x_{n} &lt; \epsilon.
 
If g(x) --> A when x --> a then p(g(x)) --> p(A) when x --> a.
 
  • #10
Hmm... what if p was the square root function and A was negative. ?
 
  • #11
Well A needs to be in the domain of p for it to make sense. Guess I should have written that..
 
  • #12
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.
 
  • #13
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
 
  • #14
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
 
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