If x_1 = 3 and x_{n+1} = sqrt{2 + x_n} for n | vedclass

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(B) Given the sequence $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$.We calculate the first few terms:$x_2 = \sqrt{2 + x_1} = \sqrt{2 + 3} = \sqrt{5} \appr

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$2$

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(B) Given the sequence $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$.We calculate the first few terms:$x_2 = \sqrt{2 + x_1} = \sqrt{2 + 3} = \sqrt{5} \approx 2.236$$x_3 = \sqrt{2 + x_2} = \sqrt{2 + \sqrt{5}} \approx \sqrt{4.236} \approx 2.058$It is observed that $x_1 > x_2 > x_3 > \dots$. The sequence is monotonically decreasing and bounded below by $2$.Since the sequence is monotonic and bounded,it must converge to a limit $L$.Let $\lim_{n 	o \infty} x_n = L$.Taking the limit on both sides of the recurrence relation $x_{n+1} = \sqrt{2 + x_n}$:$L = \sqrt{2 + L}$$L^2 = 2 + L$$L^2 - L - 2 = 0$$(L - 2)(L + 1) = 0$Since $x_n > 0$ for all $n$,the limit $L$ must be positive.Therefore,$L = 2$.

If $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$ for $n \ge 1$,then $\lim_{n \to \infty} x_n$ is equal to

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