Prove sequence ${x_{n + 1}} = \sin {x_n},{x_1} = 1 $ has a limit

Question

Prove that the sequence defined by $${x_{n + 1}} = \sin {x_n},\ {x_1} = 1$$ has a limit.

Ok, I want to prove by Weierstrass:

How can I do it?

"Sequece is limited" do you mean sequence is "bounded"? – haqnatural Aug 27 '15 at 19:36 — haqnatural, Aug 27 '15 at 19:36
Related: http://math.stackexchange.com/questions/45283 – Did Aug 27 '15 at 19:58 — Did, Aug 27 '15 at 19:58

score 4 · Accepted Answer · answered Aug 27 '15 at 19:30

4

Hint: The inequality $\sin(x) \le x$ holds for all $x \ge 0$. The second assertion is trivial.

answered Aug 27 '15 at 19:30

Dominik

2

I think you need $0\le \sin x$ for $0 \le x \le 1$ to keep your hint applicable. Saying $-1 \le \sin x \le 1$ for all $x$ is not enough. – Henry Aug 27 '15 at 19:37
I thought about it. I didn't think that $x \ge 0$ is trivial. Ok, thanks! – Evgeny Semyonov Aug 27 '15 at 19:40
@Henry It's ok if hints are just hints and not complete :) – Hagen von Eitzen Aug 27 '15 at 20:20

1 Answers1