Limit $\lim _{x \to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=1$ I've been investigating some interesting infinite square roots, and I've arrived at the hypothesis that $$\lim_{x\to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=1$$
However, I have tried to prove this but have found myself unable to do so. For example, I've tried re-writing this as $$1=\sqrt{x+1}$$ so $1=x+1$, which leads us to $x=0$, which doesn't exactly work- replacing $x$ with $0$ yields a value of $0$.
That's another side point: obviously the method I just used yields an incorrect result, but where is the maths flawed?
Please could you either prove or disprove my hypothesis? Thank you in advance.
 A: Take $a_0 = \sqrt{x}$ and $a_{n+1} = \sqrt{x+a_n}$. We need to show that
$1$. $a_{n+1} > a_n$ (the sequence is monotonically increasing)
$2$. There exists an $m$ such that $a_n \leq m$ for all $n$ (the sequence is bounded)
$1$ is easy. We have $a_0 = \sqrt{x}$ and $a_1 = \sqrt{x+\sqrt{x}}$. First, $a_1 > a_0$ as we have
$$\sqrt{x} > 0 \iff x+\sqrt{x} > x \iff \sqrt{x+\sqrt{x}}>\sqrt{x} \iff a_1 > a_0$$
Assume $1$ holds up to $n$. Then $a_{n+1} = \sqrt{x+a_n} > \sqrt{x+a_{n-1}} = a_n$ so $1$ holds for $a_{n+1}$. By induction, $1$ holds such that $a_n$ is monotonically increasing.
Now for $2$, we do the following. You can use the common way to solve this kind of radical, which is to assign a value to it $y$:
$$y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}$$
$$y^2 = x+\sqrt{x+\sqrt{x+\ldots}}$$
$$y^2=x+y$$
$$y^2-y-x=0$$
Using the quadratic equation,
$$y=\frac{1 \pm \sqrt{1+4x}}{2}$$
$y > 0$ so the smaller solution is extraneous.
$$y = \frac{1+\sqrt{1+4x}}{2}$$
this means
$$\lim_{n \to \infty} a_n = \frac{1+\sqrt{1+4x}}{2}$$
and we can prove the bound
$$a_n \leq \frac{1+\sqrt{1+4x}}{2} $$
inductively. First, we have $a_0 = \sqrt{x} < \frac{1}{2} + \sqrt{\frac{1}{4}+x} = \frac{1+\sqrt{1+4x}}{2}$. Now assume the inequality is true for all $a_i$ for $i \leq n$. Then, for $x\geq 0$,
$$\Big(\frac{1+\sqrt{1+4x}}{2}\Big)^2 = \frac{1+2\sqrt{1+4x}+(1+4x)}{4} = x + \frac{1+\sqrt{1+4x}}{2}$$
so
$$\frac{1+\sqrt{1+4x}}{2} = \sqrt{x+\frac{1+\sqrt{1+4x}}{2}}$$
Then
$$a_{n+1} = \sqrt{x+a_n} \leq \sqrt{x+\frac{1+\sqrt{1+4x}}{2}} = \frac{1+\sqrt{1+4x}}{2}$$
Therefore, this sequence is bounded and monotonically increasing, so it converges. Now we can evaluate: at $0$, it comes out to
$$y\vert_0=\lim_{x \to 0}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}= \frac{1+
\sqrt{1}}{2}=1$$
A: This is an alternative to Ryan Shesler's answer, differing mainly in the upper bound that establishes that the limit exists.
Let $x_0=0$ and $x_{n+1}=\sqrt{x+x_n}$, with $x\gt0$. Then $x_n$ is an increasing sequence since $x_1=\sqrt x\gt0=x_0$ and $x_n\gt x_{n-1}$ implies $x_{n+1}=\sqrt{x+x_n}\gt\sqrt{x+x_{n-1}}=x_n$, and $x_n$ is bounded above by $1+x$ since $x_0=0$ is certainly less than $1+x$ and if $x_n\lt1+x$ then $x_{n+1}=\sqrt{x+x_n}\lt\sqrt{x+(1+x)}=\sqrt{1+2x}\lt1+x$. Consequently the limit as $n\to\infty$ exists.
If $L=\lim_{n\to\infty}x_n$, then $L^2=x+L$, which solves to $L=(1\pm\sqrt{1+4x})/2$, but only the positive root is possible (since $x_n\gt0$ for all $n\gt0$). Taking the limit of $(1+\sqrt{1+4x})/2$ as $x\to0^+$ gives $1$.
A: Let $\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=y\implies  \sqrt{x+y}=y$
$$\implies y^2-y-x=0$$
Solving above quadratic equation for $y>0$, we get $$y=\frac{1+\sqrt{1+4x}}{2}$$
$$\therefore \lim_{x\to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=\lim_{x\to 0}\frac{1+\sqrt{1+4x}}{2}=1$$
