Proving a lim

**thedoctor818** · January 30th 2010, 06:58 PM

I am trying to find the limit of this sequence: $s_n = \dfrac{sin (n)}{n}.$ Any help would be appreciated.

**minesweeper26** · January 30th 2010, 07:13 PM

consider |sin(n)| =< 1 for all n.

**thedoctor818** · January 30th 2010, 07:39 PM

Thanks, that helps some, but I am still not sure about what to take as N.

**minesweeper26** · January 30th 2010, 07:42 PM

y not greater than 1 over epsilon?

**lovemath** · January 30th 2010, 08:10 PM

$lim \frac{sin(n)}{n}$
Have: $\frac{sin(n)}{n} \leq \frac{1}{n}$
But $lim \frac{1}{n} = 0$
So: $lim \frac{sin(n)}{n} =0$

**xalk** · January 31st 2010, 01:58 AM

Originally Posted by thedoctor818

I am trying to find the limit of this sequence: $s_n = \dfrac{sin (n)}{n}.$ Any help would be appreciated.

For that limit you have to use the following theorem:

If $lim_{n\to\infty} x_{n} = 0$ and the sequence $y_{n}$ is bounded ,then $lim_{n\to\infty} x_{n}y_{n}=0$ .

And in our case :

$y_{n} = sin(n)$ where $|sin(n)|\leq 1$ for all n ,and

$x_{n} =\frac{1}{n}$ ,where $lim_{n\to\infty}\frac{1}{n}=0$

Hence $lim_{n\to\infty}\frac{sin(n)}{n} = 0$

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