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

1. anonymous

looks as though that diverges unless there is supposed to be a division in that formula

2. anonymous

using the product rule for limits you can first evaluate the limit of sqrt(n) and then the limit in the bigger brackets

3. anonymous

1/6 seems a very peculiar answer

4. anonymous

there is nothing to stop that limit diverging

5. anonymous

$\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}-\sqrt{9n})=$ $=\lim_{n\rightarrow +\infty}\sqrt{n}\cdot (\sqrt{9n+1}-\sqrt{9n}) \cdot\frac{\sqrt{9n+1}+\sqrt{9n}}{\sqrt{9n+1}+\sqrt{9n}}=$ $=\lim_{n\rightarrow +\infty}\frac{\sqrt{n}}{\sqrt{9n+1}+\sqrt{9n}}=\lim_{n\rightarrow +\infty}\frac{1}{\sqrt{9+1/n}+\sqrt{9}}=\frac{1}{6}$

6. anonymous

thanks