Question

lim sin(n!) n approaches to infinity

Answer 1

maybe it is \(\lim_{n \to \infty } \sin(n!)/n\) ?

Answer 2

if so, then lim is equal to 0. It is because sin x is a bounded function [-1,1].

Answer 3

hi, there
I mean lim sin(n!) as n tends to infinity.

Answer 4

then there is no limit. Because sin x is a periodic function

Answer 5

hi, I need an exact proof. For instance for divergence of sin(n) we can arrive at a contradiction. But how about this?

Answer 6

The limit is zero as @myko said

Answer 7

Here is a complete proof
\[

\Large \left|\frac {\sin(n!) } { n }\right|\le \frac 1 n\to 0 \text { as } n\to \infty

\]