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 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.
