# Thread: Proving a lim

1. ## Proving a lim

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

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

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

4. y not greater than 1 over epsilon?

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

6. Originally Posted by thedoctor818
I am trying to find the limit of this sequence: $\displaystyle s_n = \dfrac{sin (n)}{n}.$ Any help would be appreciated.
For that limit you have to use the following theorem:

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

And in our case :

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

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

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