$\displaystyle \lim_{n\rightarrow \infty } (-1)^n n \sin({\pi \sqrt{n^2+\frac{2}{3}n+1}}) \sin(\frac{1}{n})$

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- Jan 2nd 2013, 07:20 AMrozuutLimit
$\displaystyle \lim_{n\rightarrow \infty } (-1)^n n \sin({\pi \sqrt{n^2+\frac{2}{3}n+1}}) \sin(\frac{1}{n})$

- Jan 2nd 2013, 03:04 PMchiroRe: Limit
Hey rozuut.

I'm not sure if you have to use delta-epsilon or not, but you can use the fact that 2*sin(theta)*cos(theta) = sin(2*theta) and show that if theta doesn't converge then neither does the limit.

You will have to also partition (-1)^n * n with the other term as well, but the above is my suggestion to showing that it doesn't exist (you can show that (-1)^n also doesn't exist by writing it as a trig term of cos(pi*n) = (-1)^n). - Jan 2nd 2013, 04:26 PMSorobanRe: Limit
Hello, rozuut!

I have some ideas to get you started . . .

Quote:

$\displaystyle \displaystyle\lim_{n\to\infty}\left[ (\text{-}1)^n\cdot n\cdot \sin\left(\pi \sqrt{n^2+\tfrac{2}{3}n+1}}\right)\cdot\sin\left( \tfrac{1}{n}\right)\right]$

We have: .$\displaystyle \lim_{n\to\infty}\left[(\text{-}1)^n\cdot \sin\left(\pi \sqrt{n^2+\tfrac{2}{3}n+1}\right) \cdot \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} \right] $

. . $\displaystyle =\;\lim_{n\to\infty}(\text{-}1)^n\cdot\lim_{n\to\infty} \sin\left(\pi \sqrt{n^2+\tfrac{2}{3}n+1}\right) \cdot \lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right )}{\frac{1}{n}} $

The first limit fluctuates between +1 and -1.

The second limit is the sine of a large multiple of $\displaystyle \pi.$

. . . . $\displaystyle \lim_{m\to\infty}\sin(m\pi)$

The third limit is 1: .$\displaystyle \lim_{n\to\infty}\frac{\sin(\frac{1}{n})}{\frac{1} {n}} \;=\;\lim_{k\to0}\frac{\sin k}{k} \;=\;1 $

Can you continue?