Find $\lim\limits_{n\to \infty}x_{n}$ if for all $n\ge 1$ $\{x_n\}$denotes a sequence of real numbers where $x_{n+1}=\frac{1}{2}(x_n+\frac{5}{x_n})$

Question

Also given

Let x be a positive number.

$x_{1}=\dfrac{1}{2}(x+\dfrac{5}{x})$, $x_{2}=\dfrac{1}{2}(x_1+\dfrac{5}{x_1})$

I have shown that for all $n\ge 1$, $\dfrac{x_n-\sqrt{5}}{x_n+\sqrt{5}}={\bigg(\dfrac{x-\sqrt{5}}{x+\sqrt{5}}}\bigg)^{2^{n}}$

But from here I get $x_n=\dfrac{\sqrt{5}\cdot 2 \cdot \bigg({\bigg(\dfrac{x-\sqrt{5}}{x+\sqrt{5}}}\bigg)^{2^{n}}+1\bigg)}{1-{\bigg(\dfrac{x-\sqrt{5}}{x+\sqrt{5}}}\bigg)^{2^{n}}}$

But from here how do I get $lim_{n\to \infty}x_{n}$

L'Hospital does not help me here as I think.

It is a GREAT GRAND SHAME ON MYSELF that I could not think of the problem in a simpler way.

I should edit this:

We know

$x_{n}=\dfrac{1}{2}(x_{n-1}+\dfrac{5}{x_{n-1}})$

$\implies \dfrac{1}{2}(x_{n-1}+\dfrac{5}{x_{n-1}})\ge \bigg(x_{n-1}\cdot \dfrac{5}{x_{n-1}} \bigg)^{1/2}$[By the A.M-G.M inequality]

$\implies x_n \ge \sqrt{5}$

Thus $\lim\limits_{n\to \infty}x_{n}= \sqrt{5}$

Done!!I should die right now!!

Answer 1

By induction we get easy that all $$x_i>0$$, so we Can use the $AM-GM$ inequality and we get $$\frac{1}{2}\left(x_n+\frac{5}{x_n}\right)\geq \sqrt{x_n\cdot \frac{5}{x_n}}=\sqrt{5}$$

Answer 2

Given $ \ x_{n+1}=\frac{1}{2} (x_n+\frac{5}{x_n} ) \\ $ ,.........(1)

Let $ \ \lim_{n \to \infty} x_n=l \ $ , then $ \ \lim_{n \to \infty} x_{n+1}=l \ $

Hence taking limit $ \ n \to \infty \ $ of both sides of $ \ (1) \ $ , we get

$$ l=\frac{1}{2} (l+\frac{5}{l}) \\ \Rightarrow 2l^2=l^2+5 \\ \Rightarrow l^2=5 \\ \Rightarrow l=\sqrt 5 $$

Therefore,

$$ \lim_{n \to \infty} x_n=l \ =\sqrt 5 $$