Conjugation on limit

The fact is, for any positive integer $n$ larger than 2, we have $\lim _{ x\to \infty } \left( \underbrace { \sqrt { x+\sqrt { x+\sqrt { x+...\sqrt { x } } } } }_{ n } -\sqrt { x } \right) =\frac { 1 }{ 2 }.$ Before proving this limit, we need the following lemma.

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Lemma. For any real number $a>0,$ there exists a real number $x$ such that if $N\geq x$ , then $\sqrt{N+a}>a.$

Proof. This follows from the fact that $\sqrt{x+a}$ is a strictly increasing function.

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With that in mind, we can proceed with the evaluation of the limit. We first consider the case when $n$ tends to infinity. Then, we have $\begin{aligned} u&:=& \sqrt { x+\sqrt { x+\sqrt { x+... } } }\\ &=&\sqrt{x+u}\\ u^2-u-x&=&0\\ u&=&\frac{1+\sqrt{1+4x}}{2} \end{aligned}$ since $u>0.$ Thus, the limit evaluates to $\begin{aligned} \lim_{x\to\infty}(u-\sqrt{x})&=&\lim_{x\to\infty}\left(\frac{1+\sqrt{1+4x}}{2}-\sqrt{x}\right)\\ &=&\lim_{x\to\infty}\left(\frac{1+\sqrt{1+4x}-2\sqrt{x}}{2}\right)\\ &=&\lim_{x\to\infty}\left(\frac{1+\sqrt{1+4x}-\sqrt{4x}}{2}\right)\\ &=&\frac{1}{2}. \end{aligned}$ Now, using the lemma above, we know that for any positive integer $n,$ the following inequality holds for $x>0.$ $u\geq\underbrace { \sqrt { x+\sqrt { x+\sqrt { x+...\sqrt { x } } } } }_{ n }\geq\sqrt{x+\sqrt{x}}.$ Hence, the result follows directly from squeeze theorem.

If you use series expansions of these radicals then each of them starts with 1/2 followed by other terms which -> 0 as x -> infinity