the limit as n -> infinity sqrt(n)[(sqrt(9n+1)-sqrt(9n)].

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the limit as n -> infinity sqrt(n)[(sqrt(9n+1)-sqrt(9n)].

Mathematics
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looks as though that diverges unless there is supposed to be a division in that formula
using the product rule for limits you can first evaluate the limit of sqrt(n) and then the limit in the bigger brackets
1/6 seems a very peculiar answer

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there is nothing to stop that limit diverging
\[\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}-\sqrt{9n})=\] \[=\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}-\sqrt{9n}) \cdot\frac{\sqrt{9n+1}+\sqrt{9n}}{\sqrt{9n+1}+\sqrt{9n}}=\] \[=\lim_{n\rightarrow +\infty}\frac{\sqrt{n}}{\sqrt{9n+1}+\sqrt{9n}}=\lim_{n\rightarrow +\infty}\frac{1}{\sqrt{9+1/n}+\sqrt{9}}=\frac{1}{6}\]
thanks

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