find the limit of the following sequence or determine that the sequence diverges. {sin n / 2n} select the…

Explanation:

Step1: Recall the range of sine function

We know that $- 1\leqslant\sin n\leqslant1$ for all real - valued $n$. So, $-\frac{1}{2n}\leqslant\frac{\sin n}{2n}\leqslant\frac{1}{2n}$.

Step2: Find the limits of the bounding sequences

We find $\lim_{n\rightarrow\infty}-\frac{1}{2n}=0$ and $\lim_{n\rightarrow\infty}\frac{1}{2n}=0$.

Step3: Apply the Squeeze Theorem

By the Squeeze Theorem, if $a_n\leqslant b_n\leqslant c_n$ for all $n$ greater than some positive integer $N$ and $\lim_{n\rightarrow\infty}a_n = L$ and $\lim_{n\rightarrow\infty}c_n = L$, then $\lim_{n\rightarrow\infty}b_n = L$. Here, $a_n=-\frac{1}{2n}$, $b_n = \frac{\sin n}{2n}$, $c_n=\frac{1}{2n}$, and $L = 0$. So, $\lim_{n\rightarrow\infty}\frac{\sin n}{2n}=0$.

Answer:

A. The limit of the sequence is $0$