# Calculating limits

1. Nov 16, 2011

### Ted123

Show that, with $t\in [0,1]$ : $$\lim_{n\to\infty} \frac{\sin(nt)}{n}=0$$

This is easy but I've forgotten how to calculate limits. Can anyone jog my memory?

Last edited: Nov 16, 2011
2. Nov 16, 2011

### MaxManus

Not sure, but how can the limit be something else when |sin| <=1 and n goes to infinity?