First slide

Introduction to limits

Question

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

Moderate

A

-1
B

2
C

$\sqrt{5}$
D

3

Solution

$\begin{array}{l} = x_{n + 1} \sqrt{2 + x_{n}} \\ or \lim_{n \to \infty} x_{n + 1} = \sqrt{2 + \lim_{n \to \infty} x_{n}} \\ or t = \sqrt{2 + t} (\lim_{n \to \infty} x_{n + 1} = t) \\ or t^{2} - t - 2 = 0 \\ or (t - 2) (t + 1) = 0 \\ or t = 2 (x_{n} > 0 \forall n, t > 0) \end{array}$

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