if a_(n+1) = (√(a_1 + a_n )) find lim_(x→∞) a


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Question Number 147529 by mathdanisur last updated on 21/Jul/21

$i f a_{n + 1} = \sqrt{a_{1} + a_{n}}$ $Double subscripts: use braces to clarify$
Commented by mathdanisur last updated on 21/Jul/21

$S o r r y a_{1} = 12$
	Answered by gsk2684 last updated on 21/Jul/21

	$L = \lim_{n \to \infty} a_{n + 1} = \lim_{n \to \infty} \sqrt{a_{1} + a_{n}} > 0$ $L = \sqrt{a_{1} + L}$ $L^{2} - L - a_{1} = 0$ $L = \frac{1 \pm \sqrt{1 + 4 a_{1}}}{2} > 0$ $L = \frac{1 + \sqrt{1 + 4 a_{1}}}{2} = \frac{1 + \sqrt{1 + 48}}{2} = 4$
	Commented by Snail last updated on 21/Jul/21

	$Y o u a r e t e c h n i c a l l y w r o n g i n t h e s e c o n d l i n e$ $w h i l e w r i t i n g L u h a v d t o s h o w a t f i r s t t h e s e q$ $i s e i t h e r m o n o t o n i c i n c r e a s i n g o r d e c r e a s i n g$
	Answered by Snail last updated on 21/Jul/21

	$C l a i m a_{n + 1} ⩾ a_{n}$ $G o t h r o u g h b a s i c s t e p s o f i n d u c t i o n$ $a n d t h e n$ $F o r (m + 1) c a s e$ $w e h a v e a_{m + 1} ⩾ a_{m}$ $\sqrt{(} a_{1} + a_{m + 1}) ⩾ \sqrt{(} a_{1} + a_{m})$ $a_{m + 2} ⩾ a_{m} h e n c e t h e s e q i s m o n o t o n i c i n c r e a s$ $N o w u c a n d o a s l i k e i t i s d o n e$
	Commented by gsk2684 last updated on 21/Jul/21

	$y e s, t h a n k y o u$
	Answered by mathmax by abdo last updated on 21/Jul/21

	$a_{n + 1} = f (a_{n}) with f (x) = \sqrt{x + 12} (a_{1} = 12)$ $f is defined continue on [- 12, + \infty [$ $f^{^{'}} (x) = \frac{1}{2 \sqrt{x + 12}} > 0 \Rightarrow f is increazing on] - 12, + \infty [$ $so \lim_{n \to + \infty} a_{n} = x_{0} / f (x_{0}) = x_{0} (fix point of f)$ $\Rightarrow x_{0} = \sqrt{x_{0} + 12} \Rightarrow x_{0}^{2} - x_{0} - 12 = 0$ $Δ = 1 + 4 (12) = 49 \Rightarrow x_{1} = \frac{1 + 7}{2} = 4 and x_{2} = \frac{1 - 7}{2} = - 3 < 0$ $let prove [that a_{n} > 0 \forall n$ $a_{2} = \sqrt{a_{1} + 12} > 0 (true) let suppose a_{n} > 0 \Rightarrow$ $a_{n + 1} = \sqrt{a_{n} + 12} > 0 \Rightarrow \lim_{n \to + \infty} a_{n} = 4$
	Commented by mathdanisur last updated on 21/Jul/21

	$t h a n k y o u S i r$
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