# epsilon delta proof

• February 7th 2011, 05:57 PM
Noxide
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?
• February 7th 2011, 06:03 PM
roninpro
I'm not very clear on what you are asking. Are you trying to show

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

using the $\varepsilon$ definition?
• February 7th 2011, 06:21 PM
Prove It
Use $\displaystyle \left|\frac{\sin{n}}{n}\right| \leq \left|\frac{1}{n}\right|$