Q. 55 - Calculus

Question

Evaluate the given limits.

  

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

Step-by-Step Solution

Answer

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

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,

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

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