<s>[INST] 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.
 [/INST] 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*}
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