Question

Evaluate the given limits.

$\underset{i\to \mathrm{\infty }}{lim} \frac{\sin k}{2^k}\text{(Hint: Use the Squeeze Theorem.)}$

Short answer

Ans: The limit of the sequence$\left\{a_k\right\}=\frac{\sin k}{2^k}$is$0$.

Step-by-step solution

Step 1. Given information.

given,

$\underset{i\to \mathrm{\infty }}{lim} \frac{\sin k}{2^k}\text{(Hint: Use the Squeeze Theorem.)}$

Step 2. The objective is to determine whether the sequence is convergent or divergent and to find the limit of the sequence if the sequence is convergent. 

In the sequence $\left\{a_k\right\}=\frac{\sin k}{2^k}$the general term is $a_k=\frac{\sin k}{2^k}$

The sin function is bounded and lies between

$-1\le \sin k\le 1$

for$k>0,2^k>0$, therefore,

$\frac{-1}{2^k}\le \frac{\sin k}{2^k}\le \frac{1}{2^k}$

Therefore, the given sequence is bounded.

Step 3. Now,

The sequences $\left\{\frac{-1}{2^4}\right\}$and $\left\{\frac{1}{2^4}\right\}$are geometric sequences with a common ratio of less than Therefore, the sequences $\left\{\frac{-1}{2^4}\right\}$and $\left\{\frac{1}{2^4}\right\}$ are convergent and converge to $0$.

Step 4. Now,

By Squeeze Theorem, the limit of the function $\frac{\sin k}{2^k}$is 0 as the limit of $\left\{\frac{-1}{2^4}\right\}$and $\left\{\frac{1}{2^4}\right\}$is $0$.

Therefore,

role="math" localid="1649304009851" $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left\{a_k\right\}=\underset{k\to \mathrm{\infty }}{lim} \left\{\frac{\sin k}{2^k}\right\}\\ =0\end{array}$

Thus the limit of the sequence$\left\{a_k\right\}=\frac{\sin k}{2^k}$ is$0$.