Answer
$$0$$
Work Step by Step
Squeeze Theorem for sequence: Let us consider that $\{a_n\}$,$\{b_n\}$ and $\{c_n\}$ are sequences for a some number $K$.
That is, $b_n \leq a_n \leq c_n$ for $n \gt K$ and $\lim\limits_{n \to \infty} b_n= \lim\limits_{n \to \infty} c_n=l$
Here, we have $a_n=\dfrac{\sin(n)}{2^n}$
Since, $-1 \leq \sin (n) \leq 1$; for all $n$.
We will multiply $a_n$ by $\dfrac{1}{2^n}$ to obtain:
$\dfrac{-1}{2^n} \leq \dfrac{\sin(n)}{2^n} \leq \dfrac{1}{2^n}$
As per Squeeze Theorem, we have $\{b_n\}=-\dfrac{1}{2^n}$ and $\{c_n\}=\dfrac{1}{2^n}$
Now, $\lim\limits_{n \to \infty} b_n=\lim\limits_{n \to \infty} (-\dfrac{1}{2^n})=0$
and $\lim\limits_{n \to \infty}c_n=\lim\limits_{n \to \infty} (\dfrac{1}{2^n})=0$
This implies that $\lim\limits_{n \to \infty} a_n=0$ by Squeeze Theorem for sequence.
