Question

2. Assuming that the sequence \{$a_n\}$ converges, find the limit of the sequence defined by $a_1 = 3$, $a_{n+1} = 12 - \sqrt{a_n}$.

          2. Assuming that the sequence \{$a_n\}$ converges, find the limit of the sequence defined by
$a_1 = 3$, $a_{n+1} = 12 - \sqrt{a_n}$.
        
2. Assuming that the sequence {an} converges, find the limit of the sequence defined by
a1 = 3, an+1 = 12 - √(an).

Added by Derrick G.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Assuming that the sequence {an} converges, find the limit of the sequence defined by a1 = 3, an+1 = 12 - √an
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Transcript

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00:01 So this is our question number 106 okay this is our first term which is 3 and here n plus 1 term which is 12 minus square root of a n so let us assume that the limit of an is convert to x and also limit of a n plus 1 is also converges to x here we have a n plus 1 is equal to 12 minus square root of a n okay now apply limit okay but your limit n turns to infinite of a n plus 1 is x and here limit n tends to infinite of square root of n is also x now your x minus 12 is equal to negative x now take square both sides so here x minus 12 square is equal to x vector out x minus 12…
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