We will first prove that the sequence is strictly increasing. Note that for each $n$ we have $x_n \le \frac{\pi}{2}$, since $\sin x \le 1$ for each $x \in \mathbb{R}$. Besides we have that the function $f(x) = \sin x - x$ is decreasing for $x \in \mathbb{R}$, because $f'(x) = \cos x - 1 \le 0$ for each $x \in \mathbb{R}$.

Therefore, for $x \in (-\infty, \frac{\pi}{2})$ we have $f(x) \ge \sin \frac{\pi}{2} - \frac{\pi}{2}$. So we get that $\sin x_n \ge x_n + 1 - \frac{\pi}{2}$, which is equivalent to $x_{n+1} \ge x_n$. Therefore, the series $(x_n)_{n=1}^\infty$ is increasing and since it is bounded it follows that it is convergent. If $l$ is its limit, then for $l$ it is true that $l = \sin l + \frac{\pi}{2} - 1$, i.e. $f(l) = f(\frac{\pi}{2})$. The function $f$ is decreasing, which means that $l = \frac{\pi}{2}$.

$\boxed{\frac{\pi}{2}}$