Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$ 
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$

Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
 A: If you factor out a $\sqrt{x}$ term from the denominator one has
\begin{align*}
\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} &= \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{x} \sqrt{x + \sqrt{x}}}}\\
&= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}}}}}\\
&= 1.
\end{align*}
A: $$\text{Let}\quad x=\frac{1}{\epsilon^2} \quad\implies\quad
\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\frac{1}{1+\epsilon\:\sqrt{1+\epsilon}}\qquad\qquad \epsilon\neq 0$$
$$\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon\:\sqrt{1+\epsilon}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon\sqrt{1}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon} =1$$
A: Let $y=√x$. $\lim x \rightarrow \infty =\lim y \rightarrow \infty. $
Numerator: $y$
Denominator:
$\sqrt {y^2 +\sqrt{y^2+y}}= \sqrt{y^2+y\sqrt{1+1/y}}=$
$y\sqrt{1+(1/y)\sqrt{1+1/y}}.$
$\lim_{y \rightarrow \infty} \dfrac{y}{y \sqrt{1+(1/y) \sqrt{1+1/y}}}= $
$\lim_{y \rightarrow \infty} \dfrac{1}{\sqrt{1+(1/y)\sqrt{1+1/y}}} =1.$
A: Note that
$$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}}$$
You can also look at it as 
$$\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$$ In case dividing by $\sqrt{x}$ bothers you.
A: A fun overkill: it is well known (at least among Ramanujan supporters) that for any $x>1$ we have
$$ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \tfrac{1}{2}+\sqrt{x+\tfrac{1}{4}} $$
hence $\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$ is bounded between $1$ and $\frac{\sqrt{x}}{\sqrt{x+\frac{1}{4}}+\frac{1}{2}}$, whose limit as $x\to +\infty$ is also $1$.
The claim hence follows by squeezing.
A: Divide by $\sqrt{x}$ to get 
$$\lim_{x \to \infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac 1x + \sqrt{\frac{1}{x^3}}}}} = 1$$
A: Let $\Lambda =$ the limit we need to find. Then, $ \ \Box\ \Lambda = 1$.

Proof: We will begin our proof using the following Lemma. $$\forall a, b\in\mathbb{R}, \ \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}.\tag1$$
Substitute $a = x$ and $b = x + \sqrt{x}$ into the Lemma. $$(1) = \sqrt{\frac{x + \sqrt{x^2 - x + \sqrt{x}}}{2}} + \sqrt{\frac{x - \sqrt{x^2 - x + \sqrt{x}}}{2}}.$$ Find the limit of the fractions under each root.$$\lim_{x\to\infty}\frac{x \pm \sqrt{x^2 - x + \sqrt{x}}}{2} = \frac 12\lim_{x\to\infty}\bigg(x \pm \sqrt{x^2 - x + \sqrt{x}}\bigg) = \frac12\cdot\infty = \infty$$ $$\therefore \lim_{x\to\infty}\sqrt{x + \sqrt{x + \sqrt{x}}} = \sqrt{\infty} + \sqrt{\infty} = \infty + \infty = \infty.$$ And, $\because \lim_{x\to\infty}\sqrt{x} = \infty$ then we finally have as desired. $$\Lambda = \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x +\sqrt{x}}}} = \frac{\infty}{\infty} = 1$$ $$\therefore \Lambda = 1.\tag*{$\bigcirc$}$$
