Find the limit of n(sqrt(n+1) - sqrt(n))^2

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SUMMARY

The limit of the sequence defined by xn = n(sqrt(n+1) - sqrt(n))^2 as n approaches infinity is computed by first expanding the expression. The binomial expansion leads to the form (2n + 1) - {2 · sqrt(n+1) · sqrt(n)}, which remains indeterminate. To resolve this, multiplying by the conjugate (2n + 1) + {2 · sqrt(n+1) · sqrt(n)} allows simplification. Ultimately, applying the limit of a rational function reveals the behavior as n approaches infinity.

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Homework Statement



Compute lim n-> infinity for xn = n(sqrt(n+1) - sqrt(n))^2

Homework Equations



non (as far as i know)

The Attempt at a Solution



i tried logging it, didnt get me very far though, i had logxn -> log infinty + 2 log sqrt infitity ?

pretty stuck...
 
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Mattofix said:
Compute lim n-> infinity for xn = n(sqrt(n+1) - sqrt(n))^2

If I'm reading this right, the next thing to do would be to go ahead and multiply out the binomial square to get

(n+1) - { 2 · sqrt(n+1) · sqrt (n) } + n

= (2n + 1) - { 2 · sqrt(n+1) · sqrt (n) } .

This is still an indeterminate difference, but we know what to do with those: multiply by 1 as the ratio of the conjugate factor,

(2n + 1) + { 2 · sqrt(n+1) · sqrt (n) } ,

divided by itself. The numerator simplifies considerably. Now multiply this ratio by the factor n that was originally in front of the squared term in x_n and apply what you know about the limit of a rational function as x approaches infinity.
 
thanks :o)
 

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