
epsilon delta proof
abs[sin(n)/n] <epsilon
How do I solve for n so that I can find a delta to use in my proof that the limit of sin(n)/n is 0?
Can I just say that since sin has max and min values between 1 and 1 that 1/n < epsilon? and then use 1/epsilon as my delta?

I'm not very clear on what you are asking. Are you trying to show
$\displaystyle \displaystyle \lim_{n\to \infty} \frac{\sin n}{n}=0$
using the $\displaystyle \varepsilon$ definition?

Use $\displaystyle \displaystyle \left\frac{\sin{n}}{n}\right \leq \left\frac{1}{n}\right$