Show that the set of adherence values of the sequence $(e^{ix_n})_n$ is the unit circle. An idea please. Let $(x_n)_n$  be a sequence of $\mathbb{R}$ such that $\lim_{n\to +\infty}x_n=+\infty$ and $\lim_{n\to +\infty}x_{n + 1}-x_n = 0 $. Show that the set of adherence values of the sequence $(e^{ix_n})_n$ is the unit circle. An idea please.
 A: Let $f(x) = \exp(ix)$. $f$ satisfy:
$$\left| f(x) - f(y) \right| \leq \left| x - y \right|$$
Let $u_n = x_{n+1} - x_n$; so $\lim_{n\to\infty}u_n = 0$ and $\sum_{n=1}^{\infty} u_n = +\infty$.
Let $x\in\mathbb{R}$, and let $\varepsilon > 0$.
The set
$$S_{\varepsilon} = \{(m,k)\in\mathbb{Z}^2 \mid \left| x - m - 2\pi k\right| < \frac{\varepsilon}{2}\}$$
is infinite because $\mathbb{Z} + 2\pi \mathbb{Z}$ is dense in $\mathbb{R}$.
There exists $N_0\in\mathbb{N}$, s.t. $|u_n| < \frac{\varepsilon}{2}$ for all
$n \geq N_0$.
There exists $(m,k)\in S_{\varepsilon}$ s.t. $m - x_1 > \sum_{k=1}^{N_0} u_k$.
Let $N_1 = \inf\{n\in\mathbb{N}\mid n \geq N_0,\, \sum_{k=1}^{n} u_k \geq m - x_1\}$.
$N_1 > N_0$ and $0 < m - x_1 - \sum_{k=1}^{N_1 - 1}u_k \leq  u_N < \frac{\varepsilon}{2}$
Now
$$ \left|\exp(im) - \exp(ix_{N_1}) \right| = \left|\exp(im - ix_1) - \exp\left(i\sum_{k=1}^{N_1 - 1}u_k \right)\right| < \frac{\varepsilon}{2} $$
and
$$\left|\exp(ix) - \exp(im)\right| = \left|\exp(ix) - \exp(im + 2k\pi)\right| < \frac{\varepsilon}{2}$$
So
$$\left|\exp(ix)  - \exp(ix_{N_1})\right| < \varepsilon$$
