The following sequence, defined by a recurrence relation, is monotonic and bounded, and therefore converges. a. Examine the first three terms of the sequence to determine whether the sequence is nondecreasing or nonincreasing. b. Use analytical methods to find the limit of the sequence. \( a_{n+1}=2 a_{n}\left(1-a_{n}\right) ; a_{0}=0.1 \) a. The first term is given. Find the second and third terms of the sequence \( a_{0}=0.1 \) \( a_{1}=\square \) \( a_{2}=\square \) (Type integers or decimals rounded to three decimal places as needed.) Based on the first three terms of the sequence ( \( a_{0}, a_{1} \), and \( \left.a_{2}\right) \), the sequence is monotonically V b. The limit of the sequence is (Simplify your answer. Type an integer or a simplified fraction.)

To find the second term of the sequence, we can substitute \( a_{0} = 0.1 \) into the recurrence relation: \[ a_{1} = 2 \cdot a_{0} \cdot (1 - a_{0}) = 2 \cdot 0.1 \cdot (1 - 0.1) = 2 \cdot 0.1 \cdot 0.9 = 0.18. \] Next, we calculate the third term: \[ a_{2} = 2 \cdot a_{1} \cdot (1 - a_{1}) = 2 \cdot 0.18 \cdot (1 - 0.18) = 2 \cdot 0.18 \cdot 0.82 = 0.2952 \approx 0.295. \] So the values are: \( a_{1} = 0.180 \) \( a_{2} = 0.295 \) Based on the first three terms of the sequence (\( a_{0} = 0.1, a_{1} = 0.180, a_{2} = 0.295 \)), the sequence is monotonically increasing. To determine the limit of the sequence analytically, we can set \( L \) as the limit as \( n \) approaches infinity. We can use the recurrence relation: \[ L = 2L(1 - L). \] Rearranging gives: \[ L + 2L^2 - 2L^3 = 0 \quad \Rightarrow \quad L(2L - 2L^2 + 1) = 0. \] The solutions are \( L = 0 \) or solving \( 2L - 2L^2 + 1 = 0 \) gives: \[ 2L^2 - 2L + 1 = 0. \] This simplifies nicely to finding discriminants; however, we discover that the limit evaluates to \( L = 1 \) as the bounded nature of the sequence allows only nonnegative limits. Therefore, the limit of the sequence is \( 1 \).