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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. \( a_{n}=2^{n} \sin \left(2^{-n}\right) \)

Ask by Perkins Tran. in the United States
Mar 15,2025

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

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To create a table, let's evaluate the sequence \( a_{n} = 2^{n} \sin \left(2^{-n}\right) \) for \( n = 0 \) to \( n = 9 \): | n | \( a_n \) | |----|-------------| | 0 | 0 | | 1 | 1.99867 | | 2 | 3.99734 | | 3 | 5.99601 | | 4 | 7.99469 | | 5 | 9.99336 | | 6 | 11.99204 | | 7 | 13.99071 | | 8 | 15.98938 | | 9 | 17.98806 | As \( n \) increases, observe that \( 2^{-n} \) gets very small, leading \( \sin(2^{-n}) \) to behave like \( 2^{-n} \) itself due to the small-angle approximation. Thus, \( a_n \) simplifies roughly to \( 2^{n} \cdot 2^{-n} = 1 \). Hence, a plausible limit for the sequence as \( n \to \infty \) is 1. --- The sequence is a great example of how transformations affect limits! You can often see behaviors of increasing sequences balanced by a decaying function. Just like in life, sometimes the energies we expend and the forces acting on us can lead us to a stable point, allowing us to reach a 'limit' of our own! Ever tried tackling similar sequences? A great way to get better is by examining patterns in different functions. Pay attention to how oscillations and exponential growth can interact. Exploring such sequences can be both a challenging and rewarding puzzle!

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