Show that $x_n \to x$ as $n \to \infty.$ 
Let $\{x_n\}_{n \geq 1}$ be a sequence of real numbers such that $\sin (t x_n) \to \sin (t x)$ for all $t \in \mathbb R.$ Then show that $x_n \to x.$

My aim is to take the power series of cosine and then differentiate it term by term which respects convergence. But for that we need to first show that $\{x_n\}_{n \geq 1}$ is bounded. But I am having hard time showing this. Could anyone please help me?
Thanks a bunch!
 A: Assume that the sequence $(x_n)$ is unbounded. Set
$$
A_n:=\{t \in \mathbb{R} \mid \forall m \ge n: ~ |\sin(tx_m)-\sin(tx)| \le \frac{1}{4} \}.
$$
Then $A_n$ is closed for each $n \in \mathbb{N}$ and since $\sin(tx_n)\to \sin(tx)$ $(n \to \infty)$ for each $t$, we get
$$
\bigcup_{n \in \mathbb{N}} A_n = \mathbb{R}.
$$
According to Baire's Theorem there exists $n _0 \in \mathbb{N}$ such that $A_{n_0}$ has nonempty interior. Thus, there is an interval $[a,b] \subseteq A_{n_0}$. Choose $[a,c] \subseteq [a,b]$ such that
$$
\forall t \in [a,c]: ~ |\sin(tx)-\sin(ax)| \le \frac{1}{4}. 
$$
Now we have that for all $t \in [a,c]$ and all $m \ge n_0$:
$$
|\sin(tx_m)-\sin(ax)| \le |\sin(tx_m)-\sin(tx)|+ |\sin(tx)-\sin(ax)| \le \frac{1}{2}.
$$
Since $(x_n)$ is unbounded we can choose $m_0 \ge n_0$ such that
the length of the interval $x_{m_0} \cdot [a,c]$ is $\ge 2\pi$, so
$\sin(tx_{m_0})$ with $t \in [a,c]$ can take any value in $[-1,1]$. Thus there is some $t_0 \in [a,c]$ such that
$$
1 \le |\sin(t_0x_{m_0})-\sin(ax)| \le \frac{1}{2},
$$
a contradiction. Thus $(x_n)$ is bounded.
From here it is sufficient to show that $(x_n)$ has a unique limit point: Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$ with limit $\alpha$, say. Then $\sin(tx) = \sin(t \alpha)$ $(t \in\mathbb{R})$. Taking the derivative in $t$ we get $x\cos(tx) = \alpha \cos(t \alpha)$ $(t \in\mathbb{R})$, and for $t=0$ we get $x = \alpha$. Thus $x_n \to x$ $(n \to \infty)$.
