Finding $\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$

Question

Finding

$$\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$$

Answer 1

let $f_n(x)=\underbrace{\sqrt{x-\sqrt{x-\sqrt{x-\cdots}}}}_{n}$, then $f_{n+1}(x)=\sqrt{x-f_n(x)}$. first prove $\lim_{n\to+\infty}f_n(x)=\frac{\sqrt{4x+1}-1}{2}$

$\forall x>1$ we have :$$f_1(x)\ge f_3(x)\ge\cdots f_{2n-1}(x)\ge \cdots \ge f_{2n}(x)\ge \cdots \ge f_4(x)\ge f_2(x)$$ this show $\forall x>1,f_n(x)$ is bounded . then $\lim f_n(x)$ has high-low limit. let: $$\limsup_{n\to+\infty}f_n(x)=a(x) , \liminf_{n\to+\infty}f(x)=b(x)$$ let $t_k\in N^+$ satisfy $\lim_{t\to+\infty}f_{t_k}(x)=a$,then $\lim_{t\to+\infty}\sqrt{x-f_{t_k-1}(x)}=a$

then $\lim_{k\to+\infty}f_{t_k-1}(x)=c$,then $c=b$,if not then $c>b$ , let $u_k\in N^+$satisfy $\lim_{k\to+\infty}f_{u_k}(x)=b$ ,then $\lim_{k\to+\infty}f_{u_k+1}(x)=\sqrt{x-b}>\sqrt{x-c}=a$ , but $a$ is high limlit , acontradiction .

now we have $$a=\sqrt{x-b},b=\sqrt{x-a} \Rightarrow a-b=\frac{a-b}{\sqrt{x-a}+\sqrt{x-b}}$$

if$a\not=b$,then $\sqrt{x-a}+\sqrt{x-b}=1$,after square $1=2x-a-b+2\sqrt{(x-a)(x-b)}$, notice $a+b=1$,we have $1=x+\sqrt{(x-a)(x-b)}\ge x>1$ a contradiction. so $a=b$,then $\lim_{n\to+\infty}f_n(x)$ exist . $f_{n+1}(x)=\sqrt{x-f_n(x)}$ let $n\to+\infty$ get:$$f(x)=\sqrt{x-f(x)}\Rightarrow f(x)=\frac{\sqrt{4x+1}-1}{2}$$ ($f(x)\ge0$ so it can't be $\frac{-\sqrt{4x+1}-1}{2}$) so $\lim_{x\to1}f(x)=\frac{\sqrt{5}-1}{2}$

Answer 2

Let $$ a_n=\overbrace{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-\dots}}}}}^{n\ x\text{s}}\tag{1} $$ If $x\lt1$, then $a_2^2=x-\sqrt{x}\lt0$. If $x=1$, then $a_n$ oscillates between $0$ and $1$. So we will assume that $x\gt1$.

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Since $a_n^2=x-a_{n-1}$, we have $a_n^2-a_{n-1}^2=a_{n-2}-a_{n-1}$. Therefore, $$ a_n-a_{n-1}=-\frac{a_{n-1}-a_{n-2}}{a_n+a_{n-1}}\tag{2} $$ Since $f(a)=a+\sqrt{x-a}$ is convex, $f(0)=\sqrt{x}$, and $f(x)=x$, we have that $$ f(a)\ge\sqrt{x}\tag{3} $$ Therefore, $$ a_n+a_{n-1}=f(a_{n-1})\ge\sqrt{x}\gt1\tag{4} $$ Combining $(2)$ and $(4)$, we get $$ \left|a_n-a_{n-1}\right|\le\frac{\left|a_{n-1}-a_{n-2}\right|}{\sqrt{x}}\tag{5} $$ Since $x\gt1$, the sequence $a_n$ converges.

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For a given $x\gt1$, let $$ a(x)=\lim_{n\to\infty}a_n\tag{6} $$ Then taking the limit of $$ a_n^2=x-a_{n-1}\tag{7} $$ we get that $a(x)^2+a(x)-x=0$; that is, $$ a(x)=\frac{-1+\sqrt{1+4x}}2\tag{8} $$ Therefore, $$ \lim_{x\to1^+}a(x)=\frac{-1+\sqrt5}2\tag{9} $$

Answer 3

You are defining $$ a_1(x) = \sqrt x,\qquad a_{n+1}(x) = \sqrt{x-a_n(x)} $$ This sequence converges to the $y$ solution of $$ \sqrt{x-y} = y $$ which is $$ y^2 + y - x = 0 $$ i.e. $$ f(x) = \frac{-1+\sqrt{1+4x}}{2} $$ so the limit as $x\to 1$ is $(\sqrt 5 -1)/2$.

Answer 4

Let $$y=\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-\dots}}}}$$ Then $$y^2-x=-\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-\dots}}}}$$ $$y^2-x=-y$$ $$y^2+y-x=0$$ Now solve the quadratic equation $$y=\dfrac{-1\pm\sqrt{1+4x}}2$$ For $x=1$ you have $$y_1=\dfrac{-1+\sqrt{5}}2,y_2=\dfrac{-1-\sqrt{5}}2$$ Now you can see that $y_2$ cannot be solution because $y$ cannot be negative.

Answer 5

For fixed $a\ge0 $ and $0\le x\le a$, let $f_a(x)=\sqrt{ a-x}$. Define $f_a^{\circ0}(x)=x$ and recursively $f_a^{\circ(n+1)}(x)=f_a(f_a^{\circ n}(x))$. We want to define $$ F(a)=\lim_{n\to\infty}f_a^{\circ n}(0).$$ (At least, this seems to be the most acceptable way to define the nested root expression). The basic questions are:

• For which values of $a$ are all terms $f_a^{\circ n}(0)$ defined (i.e., a real number)?
• And for which of these $a$ does the limit exist?

For $a=0$, we have $f_a^{\circ n}(0)=0$ for all $n$, hence no problem with definedness and the limit. For $0<a<1$, we have $\sqrt a>a$, hence already $f_a^{\circ 2}(0)=f_a(f_a(0))=\sqrt{a-\sqrt a}$ is not real. For $a=1$, the values of $f_a^{\circ n}(0)$ flip between $0$ and $1$, hence all terms are defined, but the limit does not exist.

Note that $f_a^{\circ 2}$ is strictly increasing on $[0,a]$. As certainly $f_a^{\circ n}(0)\le \sqrt a$ for all $n$, we conclude that $f_a^{\circ 2}(0)$ converges to some limit $G(a)$. Then $G(a)$ must be a fixed point of $f_a^{\circ 2}$, i.e., a solution of $$ (a-X^2)^2=a-X,$$ transformed: $$ \tag1(X^2+X-a)(X^2-X-a+1)=0.$$ The first factor in $(1)$ corresponds to fixed points of $f_a$, the second factor to points of order $2$. The latter are interchanged by $f_a$. But for $a>1$, one of them is negative (their product $-a+1$ is negative). Since neither $G(a)$ nor $f_a(G(a))$ can be negative, we can ignore the second factor and are left with $X^2+X-a$, which has two real roots $\frac{-1\pm\sqrt{4a+1}}{2}$. Again, one of the roots is negative, so that the only possibility is $G(a)=\frac{-1+\sqrt{4a+1}}{2}$, and, as this is a fixed point of $f_a$, also $F(a)=\frac{-1+\sqrt{4a+1}}{2}$. We conclude $$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x\ldots}}}}=\begin{cases}0&\text{for $x=0$}\\\frac{-1+\sqrt{4x+1}}{2}&\text{for $x>1$}\\\text{undefined}&\text{otherwise}\end{cases} $$ and consequently the right-sided limit at $1$ is $$\lim_{x\to 1^+}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x\ldots}}}}=\lim_{x\to1^+}\frac{-1+\sqrt{4x+1}}{2}=\frac{\sqrt 5-1}{2}.$$

Answer 6

let $y=\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$ and then $y^2+y=x$ and solve when $x=1$

$$\large y=\frac{-1+\sqrt{5}}{2}$$

Answer 7

let $f(x)=\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$
$f(x)=\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$
$f(x)=\sqrt{x-f(x)}$
$f^2(x)=x-f(x)$
$f^2(x)+f(x)-x=0$
use the quadratic formula for f(x) with a=1, b=1, c=-x
$f(x)=\frac{-1±\sqrt{1+4x}}{2}$
$\lim_{x \to 1}{\frac{-1±\sqrt{1+4x}}{2}}$
$\lim_{x \to 1}{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}}=\frac{-1±\sqrt{5}}{2}$