Question

can anyone explain how as n--->infinity sin(6/n)/sin(3/n) approach zero.
consider sequence convergent

Answer 1

sin(6/n) / sin(3/n)

Answer 2

Thanks @myko

In that case, you know that the top goes to 0 and the bottom goes to 0, so you can use L'hopitals rule. \[\lim_{n \rightarrow \infty}{\sin(6/n) \over \sin(3/n)}={\cos(6/n) \over \cos(3/n)} \cdot {-6/n^2 \over -3/n^2}=\lim_{n\rightarrow \infty}1 \cdot {2n^2 \over n^2}=2\]

Answer 3

Are you sure this is supposed to go to 0?

Answer 4

yes it should be 2