Determine whether the sequence defined by $a _ { n } = \frac { \sin 2 n } { 9 n }$ converges or diverges. If it converges, find its limit.
A) Diverges
B) 1
C) $\frac { 2 } { 9 }$
D) 0

Question

Quizplus · Accepted Answer

We will use the Squeeze Theorem to determine the limit, which states that if $a_n \leq b_n \leq c_n$ for all $n$ and $\lim_{nightarrow \infty} a_n = \lim_{nightarrow \infty} c_n = L$, then $\lim_{nightarrow \infty} b_n = L$.

Notice that $- \frac{1}{9n} \leq \frac{\sin(2n)}{9n} \leq \frac{1}{9n}$. This is because $-1 \leq \sin(2n) \leq 1$ for all $n$.

Taking the limits of the left-hand and right-hand sides of the above inequality, we get:

$$
\lim_{nightarrow \infty} \left(-\frac{1}{9n} ight) = 0 \quad 	ext{and} \quad \lim_{nightarrow \infty} \left(\frac{1}{9n} ight) = 0.
$$

By the Squeeze Theorem, we conclude that:

$$
\lim_{nightarrow \infty} \frac{\sin(2n)}{9n} = \boxed{	extbf{(D) }0}.
$$

Determine Whether the Sequence Defined By an=sin⁡2n9na _ { n } = \frac { \sin 2 n } { 9 n }an​=9nsin2n​

Determine Whether the Sequence Defined By $a _ { n } = \frac { \sin 2 n } { 9 n }$