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Question

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


A
1
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B
2
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C
5
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D
3
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Solution

The correct option is D $$2$$
$$x_{n+1} = \sqrt{2+x_n}\\$$
$$\mbox{or }\displaystyle\lim_{n\to\infty}x_{n+1} = \sqrt{2+\displaystyle\lim_{n\to\infty}x_n}\\$$
$$\mbox{or }t=\sqrt{2+t}\quad\left(\because\displaystyle\lim_{n\to\infty}x_{n+1}= \displaystyle\lim_{n\to\infty}x_{n}=t\right)\\$$
$$\mbox{or }t^2-t-2=0\\$$
$$\mbox{or }(t-2)(t+1)=0\\$$
$$\mbox{or }t=2 \quad(\because x_n>0 \forall n, t>0)$$
Hence,$$\displaystyle \lim_{n\to\infty}x_{n}=2$$

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