Analysing The "Squeeze or sandwich Theorem" of sequences.

Still on the topic of sequences, i want to discuss on a property of the limits of sequences,which is a very important theorem known as the "Sandwich or Squeeze Theorem", in one of my topics under continuity of functions i used the sandwich theorem to evaluate a function,meaning the sandwich theorem has variety of applications,we can use it in functions and sequences

Explaining The Squeeze Theorem
before i proceed to stating the theorem i will give a brief explanation of what is meant by the squeeze theorem.
This theorem says given an unknown sequence,the sequence can be sandwiched between two convergent sequences,hence allowing us to conclude that the sequence converges.

Squeeze Theorem: Suppose that $\{s_n\}$ and $\{t_n\}$ are convergent sequences, that
\begin{align*}
&lim_{n\rightarrow\infty}s_n=lim_{n\rightarrow\infty}t_n\\
&\text{and that}\\
&s_n\leq{x_n} \leq{t_n},\text{for all n}\\
\text{then}  \{x_n\}  \text{is convergent and}\\
&lim_{n\rightarrow\infty}x_n=lim_{n\rightarrow\infty}s_n=lim_{n\rightarrow\infty}t_n
\end{align*}

Proof: Let L be the limit of the two sequences. Choose $N_1$ so that $|s_n-L|<\epsilon$ if $n\geq{N_1}$ and also choose $N_2$ so that $|t_n-L|<\epsilon$ if $n\geq{N_2}$.Set $N=max\{N_1,N_2\}$. Note that $s_n-L\leq{x_n-L}\leq{t_n-L}$  for all $n$  and so $-\epsilon<s_n-L\leq{x_n-L}\leq{t_n-L}<epsilon$ from this we see that $\epsilon<x_n-L<\epsilon$ and hence we have $|x_n-L|<\epsilon$ which proves the statement of the proof.
Please note that the theorem and proof of this sequences was gotten from M.Bruckner's classical real analysis, so you can refer to the book in case you don't understand it here fully.

Example:Let $\theta$ be some real number and consider the limit $\lim_{n\rightarrow\infty}\frac{\sin{n}\theta}{n}$

Proof: the given limit is a little bit difficult to solve but at the same might seem simple to others,to evaluate this look at $\sin{n}\theta$ you will observe that none of it values lies outside the interval $[-1,1]$. Hence
\begin{align*}
-\frac{1}{n}\leq{\frac{\sin{n}\theta}{n}}\leq{\frac{1}{n}}
\end{align*}
Now notice that the two outer sequences converge to the same value 0 and so the inside sequence (the "squeezed" one)
must converge to 0 as well, why? because taking the limits will definitely converge to zero.