Chapter 7: Q. 55 (page 604)

Evaluate the given limits.
$lim_{i \to \infty} \frac{\sin k}{2^{k}} (Hint: Use the Squeeze Theorem.)$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

given,

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

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 ${a_{k}} = \frac{\sin k}{2^{k}}$ the general term is $a_{k} = \frac{\sin k}{2^{k}}$

The sin function is bounded and lies between

$- 1 \leq \sin k \leq 1$

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

\frac{- 1}{2^{k}} \leq \frac{\sin k}{2^{k}} \leq \frac{1}{2^{k}}

Therefore, the given sequence is bounded.

Step 3. Now,

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

Therefore,

role="math" localid="1649304009851" $\begin{aligned} lim_{k \to \infty} \{a_{k}\} = lim_{k \to \infty} \{\frac{\sin k}{2^{k}}\} \\ = 0 \end{aligned}$

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

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Evaluate the given limits.
$lim_{i \to \infty} \frac{\sin k}{2^{k}} (Hint: Use the Squeeze Theorem.)$

Short Answer

Step by step solution

Step 1. Given information.

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.

Step 3. Now,

Step 4. Now,

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Short Answer

Step by step solution

Step 1. Given information.

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Step 3. Now,

Step 4. Now,

One App. One Place for Learning.

Most popular questions from this chapter

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