Limits On An Infinite Nest

$It\quad can\quad be\quad done\quad much\quad more\quad simply\\ \\ y=\sqrt { x+\sqrt { x+\sqrt { x+........... } } } \\ \therefore { y }^{ 2 }=x+\sqrt { x+\sqrt { x+\sqrt { x+........... } } } \\ \therefore { y }^{ 2 }=x+y\\ \therefore { y }^{ 2 }={ 0 }^{ + }+y\\ So\quad the\quad square\quad of\quad y\quad is\quad a\quad bit\quad greater\quad than\quad y,\quad if\quad we\\ change\quad it\quad a\quad bit.\\ If\quad y\quad is\quad very\quad close\quad to\quad zero,\quad { y }^{ 2 }\quad will\quad be\quad less\quad than\quad y.\\ So\quad only\quad solution\quad is\quad \boxed { 1 } \quad \\$

TL;DR: Prove that the expression is well-defined, giving a closed form, and find the limit of that form.

First, we will prove that $\sqrt{x + \sqrt{x + \ldots}}$ exists for all positive $x$ .

To see this, let $(a_n)$ be the sequence defined by $a_1 = \sqrt{x}, a_{n+1} = \sqrt{x + a_n}$ for all natural $n$ . Then $L = \sqrt{x + \sqrt{x + \ldots}}$ is defined as $\displaystyle\lim_{n \to \infty} a_n$ .

Now, if we know that the limit $L$ exists, we know that it satisfies $L = \sqrt{x + L}$ and thus $L = \dfrac{1+\sqrt{4x+1}}{2}$ by simple solving of a quadratic equation. We will use this result as motivation: we will prove that $a_n < \dfrac{1+\sqrt{4x+1}}{2}$ for all $n$ , and that $a_n$ is increasing, thus proving that $a_n$ has a limit (since it's increasing but bounded above ).

First, we prove that it's bounded above by $\dfrac{1+\sqrt{4x+1}}{2}$ by induction. At first, this is trivially true. $a_1 = \sqrt{x} < \dfrac{1+\sqrt{4x+1}}{2}$ is true, since it simplifies to $\sqrt{4x+1} - \sqrt{4x} < 1$ which is true for positive $x$ . Now suppose that the claim is true for some $n=k$ , and we want to show that it's also true for $n=k+1$ .

Observe that $a_{k+1} = \sqrt{x + a_k}$ . From the induction hypothesis, we know that $a_k < \dfrac{1+\sqrt{4x+1}}{2}$ , and since $\sqrt{x + a_k}$ is a monotonically increasing function on $a_k$ , we have $\sqrt{x + a_k} < \sqrt{x + \dfrac{1+\sqrt{4x+1}}{2}}$ . Now, we observe the latter:

$\sqrt{x + \dfrac{1+\sqrt{4x+1}}{2}}$

$= \frac{1}{2} \cdot \sqrt{4x + 2(1+\sqrt{4x+1})}$

$= \frac{1}{2} \cdot \sqrt{4x+2 + 2\sqrt{4x+1}}$

$= \frac{1}{2} \cdot (1 + \sqrt{4x+1})$

$= \dfrac{1 + \sqrt{4x+1}}{2}$

Thus $a_{k+1} < \sqrt{x + \dfrac{1+\sqrt{4x+1}}{2}} = \dfrac{1 + \sqrt{4x+1}}{2}$ , proving the induction claim. And thus by mathematical induction, $a_n$ is bounded above by $\dfrac{1 + \sqrt{4x+1}}{2}$ .

Now, we prove that it is increasing; that is, $a_n < a_{n+1}$ for all natural $n$ . To see this, we assume for the sake of contradiction that $a_n \ge a_{n+1}$ for some $n$ and derive a contradiction.

Note that $a_{n+1} = \sqrt{x + a_n}$ . Since $a_n \ge \sqrt{x + a_n} \ge 0$ , we can square both sides to give $a_n^2 \ge x + a_n$ , or $a_n^2 - a_n \ge x$ . This simplifies to $a_n \le \dfrac{1 - \sqrt{4x+1}}{2}$ or $a_n \ge \dfrac{1 +\sqrt{4x+1}}{2}$ . But this is impossible: $a_n \le \dfrac{1 - \sqrt{4x+1}}{2}$ cannot be true as the right hand side is less than $0$ , while $a_n$ is the square root of some real number and hence not less than $0$ . Meanwhile, $a_n \ge \dfrac{1 +\sqrt{4x+1}}{2}$ is impossible from our result above. So our assumption was invalid; so $a_n < a_{n+1}$ for all natural $n$ indeed.

Thus $a_n$ is bounded above and increasing, and hence has a limit, which we have computed above as $\dfrac{1 +\sqrt{4x+1}}{2}$ . We can now substitute this to the expression:

$\lim_{x \to 0^+} \sqrt{x + \sqrt{x + \ldots}} = \lim_{x \to 0^+} \dfrac{1 +\sqrt{4x+1}}{2}$

The latter is straightforward computation. For positive $x$ , the function is continuous, thus the limit is found by substituting the value $x = 0$ to the expression: $\dfrac{1 +\sqrt{4 \cdot 0+1}}{2} = \boxed{1}$ .

Let root(x+ root(x +root(x...... =y

So, squaring on both sides,

x + root(x +root(x ..... =(y^2)

=> (y^2) - y -x=0

As x --> 0+ ,

(y^2) - y ~ 0

So, y = 0 , 1

As x --> 0+ ,

root(x) --> 0+

so, x + root(x) --> 0+, but != 0

Therefore y=1 !!!

If we equate the limit to ${y}$ we get $y^{2}$ = ${x}$ + ${y}$ . Letting ${x}$ = 0 gives ${y}$ ( ${y}$ -1) = 0, so y=0 or 1. Now we have to use a reasoned argument as to which value to use. Well as we get close to 0, $\sqrt{x}$ > ${x}$ , so as we continue the square roots the expression becomes bigger, but always stays close to 1. However, there is a better, more algebraic way of looking at it. Let ${x}$ = 0.1 to give: $y^{2}$ - ${y}$ - 0.1 = 0, so $y_{1}$ ~1.09. Now let ${x}$ = 0.01, which gives $y_{2}$ ~1.001. Clearly ${y}$ --> 1, as ${x}$ --> $\infty$
Is this what you were looking for Calvin?