Consider the formula for the following sequence. Using a calculator in radians mode, make a table with at least 10 terms and determine a plausible value for the limit of the sequence or state that it does not exist. $$ a_{n}=2^{n} \sin \left(2^{-n}\right) $$

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1. Recognize the given sequence:
   
   $$
   a_{n} = 2^{n} \sin\left(2^{-n}\right)
   $$

2. For large values of $$ n $$, the angle $$ 2^{-n} $$ is very small, so we can use the small-angle approximation:
   
   $$
   \sin\left(2^{-n}\right) \approx 2^{-n}
   $$
   
   Substituting this approximation into the expression for $$ a_n $$ gives:
   
   $$
   a_{n} \approx 2^{n} \cdot 2^{-n} = 1
   $$
   
   This suggests that as $$ n $$ increases, $$ a_n $$ should approach 1.

3. To confirm this behavior, we can calculate at least 10 terms of the sequence using a calculator (in radians mode):

| $$ n $$ | $$ 2^{-n} $$     | $$ \sin\left(2^{-n}\right) $$ | $$ a_{n} = 2^n \sin\left(2^{-n}\right) $$ |
   |---------|------------------|-------------------------------|-------------------------------------------|
   | 1       | 0.5              | 0.4794                        | $$ 2 \times 0.4794 \approx 0.9589 $$        |
   | 2       | 0.25             | 0.2474                        | $$ 4 \times 0.2474 \approx 0.9896 $$        |
   | 3       | 0.125            | 0.1247                        | $$ 8 \times 0.1247 \approx 0.9974 $$        |
   | 4       | 0.0625           | 0.0625                        | $$ 16 \times 0.0625 \approx 1.0000 $$       |
   | 5       | 0.03125          | 0.03125                       | $$ 32 \times 0.03125 \approx 1.0000 $$      |
   | 6       | 0.015625         | 0.015625                      | $$ 64 \times 0.015625 \approx 1.0000 $$     |
   | 7       | 0.0078125        | 0.0078125                     | $$ 128 \times 0.0078125 \approx 1.0000 $$   |
   | 8       | 0.00390625       | 0.00390625                    | $$ 256 \times 0.00390625 \approx 1.0000 $$  |
   | 9       | 0.001953125      | 0.001953125                   | $$ 512 \times 0.001953125 \approx 1.0000 $$ |
   | 10      | 0.0009765625     | 0.0009765625                  | $$ 1024 \times 0.0009765625 \approx 1.0000 $$|

4. From the table, we observe that as $$ n $$ increases, the value of $$ a_n $$ approaches 1.

5. Therefore, the plausible value for the limit of the sequence is:

$$
\lim_{n \to \infty} a_{n} = 1.
   $$
As $$ n $$ increases, $$ a_n = 2^n \sin(2^{-n}) $$ approaches 1. So, the limit of the sequence is 1.

Consider the formula for the following sequence. Using a calculator in radians mode, make a table with at least 10 terms and determine a plausible value for the limit of the sequence or state that it does not exist. \( a_{n}=2^{n} \sin \left(2^{-n}\right) \)

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