$\forall t \in \mathbb{R}, e^{i \alpha_n t} \rightarrow 1 \implies \alpha_n \rightarrow 0$? $\alpha_n \in \mathbb{R}$ is a fixed sequence of real numbers, for which the following holds:
$$
\forall t \in \mathbb{R}, \lim_{n \rightarrow \infty}e^{i\alpha_nt} = 1
$$
Is it necessarily the case that $\lim_{n \rightarrow \infty} \alpha_n = 0$?
 A: Assume (as we may) that $\alpha_n\neq 0$ for all $n$. By the dominated convergence theorem and the assumption on $\alpha_n$, we have 
$$\lim_{n\to\infty} \int_0^1 e^{i\alpha_n t}dt=1\, .$$
On the other hand,
$$\int_0^1e^{i\alpha_n t}dt=\frac{1}{i\alpha_n} (e^{i\alpha_n}-1)\, .$$
Since $e^{i\alpha_n}-1\to 0$, it follows that $\alpha_n\to 0$.
A: The convergence of $e^{i\alpha_n t}$ to $1$ implies the convergence of each component (real and imaginary part):
$$\lim_{n \rightarrow \infty}sin(\alpha_nt)=0$$
$$\lim_{n \rightarrow \infty}cos(\alpha_nt)=1$$
If we take $\alpha$ as the limit of $\alpha_n$ distinct of $0$ and take $t=\frac{1}{\alpha}$ by continuity of the functions cos and sin the limit have to be $cos(1)$ and $sin(1)$ then we have a contradiction with the original statement. And then we have two options: or $\alpha_n$ converges to $0$ or not converges.
If $\alpha_n$ have a convergent subsequence for each one of they we could apply the same argument an we have that limit points are $0$ then the problem is reducted to demostrate a contradiction in having a not convergent subsequence. 
If the subsequence $\alpha_{n_k} \rightarrow \infty$ like @GEdgar said is a contradiction with the convergence of sin or cos (because the limit at infinity of both functions is not defined at infinity). Finally by this question the subsequence not having cluster points and being bounded converges to infinity, again a contradiction.
A: Let $\alpha_n$ be a sequence such that
$$
\lim_{n \to \infty}e^{i \alpha_n t} = 1\qquad\text{for all } t \in \mathbb R .
\tag{1}
$$
Suppose $\alpha_n$ does not converge to zero.  Replacing some $\alpha_n$ by $-\alpha_n$ if necessary, and taking a subsequence if necessary: we may assume there is $K>0$ so that $\alpha_n \ge K$ for all $n$.  Next, we may assume $\alpha \to \infty$: if not, there is a bounded subsequence, say $\alpha_n \le L$, and then $t = \pi/L$ contradicts the convergence (1) along that subsequence.
Now, for each $N \in \mathbb N$, define
$$
E_N = \bigcap_{n=N}^\infty \left\{t \in \mathbb R \colon \left|
e^{i\alpha_n t}-1\right| \le \frac{1}{10}\right\} .
$$
The sets $E_N$ are closed, but by (1) we have
$$
\bigcup_{N \in \mathbb N} E_N = \mathbb R .
$$
By the Baire Category Theorem, some set $E_N$ is somewhere dense: there is an interval $(a,b) \subseteq E_N$, say.  But since $\alpha_n \nearrow \infty$, there is $n \ge N$ so that $\alpha_n(b-a) > 2\pi$.  This means
$$
\{e^{i \alpha_n t}\colon t \in (a,b)\}
$$
covers the entire circle, so
$$
\left|e^{i\alpha_n t}-1\right| < \frac{1}{10}
$$
fails for some $t \in E_N$.  Contradiction.
A: $\exp$ is $2\pi i$ periodically, so the answer is No. (if you mean $t\rightarrow\infty$)
edit: see comments
