limit at $+\infty$ of x(t) for $x'(t)=\sin(x(t))$

Question

Given the initial value problem: $\left\{ \begin{array}{@{}l} x'(t) = \sin (x(t)),\ t\in\mathbb{R}\\ x(0)=x_0 \in \left]0,\pi\right[ \end{array} \right.$

I am trying to find $\lim\limits_{t \to +\infty} x(t)$

What I did:

I showed that any solution verifies $x(t) \in \left]0, \pi\right[$. I also verified that $x(t)$ is monotonic on $\mathbb{R}$, that the limit exists and that there is unique solution $x(t)$ on $\mathbb{R}$

Any help is much appreciated.

Answer 1

If the limit exists it means that $\lim_{t\to\infty} x'(t)\to0\iff \lim_{t\to\infty}\sin(x(t))\to0\iff \sin(\lim_{t\to\infty}x(t))\to0\iff \lim_{t\to\infty}x(t)\to0\lor \pi$ how can we conclude which one is the limit?

Answer 2

It is also not so hard to find the explicit solution, which is $x(t) = 2 \arctan (e^t)$ up to a time shift. Then $\lim_{t \to \infty} x(t) = \pi$ is straightforward.