anonymous
  • anonymous
the limit as n -> infinity sqrt(n)[(sqrt(9n+1)-sqrt(9n)].
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
looks as though that diverges unless there is supposed to be a division in that formula
anonymous
  • anonymous
using the product rule for limits you can first evaluate the limit of sqrt(n) and then the limit in the bigger brackets
anonymous
  • anonymous
1/6 seems a very peculiar answer

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

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