Limit $\lim _{x \to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=1$

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 $$y\vert_0=\lim_{x \to 0}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}= \frac{1+ \sqrt{1}}{2}=1$$

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 $1$.