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anonymous
 5 years ago
the limit as n > infinity sqrt(n)[(sqrt(9n+1)sqrt(9n)].
anonymous
 5 years ago
the limit as n > infinity sqrt(n)[(sqrt(9n+1)sqrt(9n)].

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0looks as though that diverges unless there is supposed to be a division in that formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using the product rule for limits you can first evaluate the limit of sqrt(n) and then the limit in the bigger brackets

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/6 seems a very peculiar answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is nothing to stop that limit diverging

nikvist
 5 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}\sqrt{9n})=\] \[=\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}\sqrt{9n}) \cdot\frac{\sqrt{9n+1}+\sqrt{9n}}{\sqrt{9n+1}+\sqrt{9n}}=\] \[=\lim_{n\rightarrow +\infty}\frac{\sqrt{n}}{\sqrt{9n+1}+\sqrt{9n}}=\lim_{n\rightarrow +\infty}\frac{1}{\sqrt{9+1/n}+\sqrt{9}}=\frac{1}{6}\]
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