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.

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. $$\Lambda = \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x +\sqrt{x}}}} = \frac{\infty}{\infty} = 1$$ $$\therefore \Lambda = 1.\tag*{$\bigcirc$}$$
