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.

Divide by $\sqrt{x}$ to get 
$$\lim_{x \to \infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac 1x + \sqrt{\frac{1}{x^3}}}}} = 1$$
