How Does the Limit of the Sequence sqrt(n)*(sqrt(n+1)-sqrt(n)) Approach 1/2?

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SUMMARY

The limit of the sequence defined by the expression sqrt(n)*(sqrt(n+1)-sqrt(n)) approaches 1/2 as n approaches infinity. This conclusion is reached through algebraic manipulation, specifically using the identity (a+b)(a-b)=a^2-b^2, which simplifies the limit to sqrt(n)/(sqrt(n+1)+sqrt(n)). An epsilon proof can further validate this limit by demonstrating that for any given ε>0, there exists an N>0 such that for all n>N, the absolute difference |a_n-1/2| is less than ε.

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tarheelborn
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Homework Statement



Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]



Homework Equations



sqrt(n)/[sqrt(n+1)+sqrt(n)] = 1/sqrt[1+(1/n)]+1

The Attempt at a Solution



I know that limit sqrt(n) = Infinity and that limit (sqrt(n+1)-sqrt(n)) = 0. And I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).

I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. I need to do an epsilon proof of the limit and I am not sure how to solve the equation in 2. in terms of epsilon. Thanks for your help.
 
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You can't find a limit using epsilon/delta.

Evaluating the limit is almost immediate after simple algebra [tex](a+b)(a-b)=a^2-b^2[/tex], and yes the limit is 1/2.

If you still want to prove it by definition you should for any given [tex]\epsilon>0\ find\ N>0\ so\ \forall\ n>N\ |a_n-1/2|<\epsilon[/tex]
 
Last edited:
tarheelborn said:

Homework Statement



Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]



Homework Equations



sqrt(n)/[sqrt(n+1)+sqrt(n)] = 1/sqrt[1+(1/n)]+1

The Attempt at a Solution



I know that limit sqrt(n) = Infinity and that limit (sqrt(n+1)-sqrt(n)) = 0. And I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).

I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. I need to do an epsilon proof of the limit and I am not sure how to solve the equation in 2. in terms of epsilon. Thanks for your help.
What do you mean "how to solve the equation in 2 in terms of epsilon"? You can get that equation using estro's hint and use it to show that the limit of the sequence is indeed equal to 1/2 as n goes to infinity.
 

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