Squeeze theorem applied to sin(n)/n

Since $\sin(n)$ always lies between $-1$ and $1$, we have $-1 \le \sin(n) \le 1$ for all positive integers $n$. Dividing every part of this inequality by $n > 0$ preserves the inequality direction, giving $-\frac{1}{n} \le \frac{\sin(n)}{n} \le \frac{1}{n}.$ This produces a lower bound sequence $-1/n$ and an upper bound sequence $1/n$, both of which sandwich $a_n = \sin(n)/n$ for every $n$.

Now evaluate the limits of the two bounding sequences: $\lim_{n\to\infty} \left(-\frac{1}{n}\right) = 0$ and $\lim_{n\to\infty} \frac{1}{n} = 0$. Since both bounds converge to the same value $0$, the Squeeze Theorem guarantees that the sequence trapped between them also converges to $0$. Therefore $\lim_{n\to\infty} \frac{\sin(n)}{n} = 0.$