<s>[INST] Calculate: $\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}$ Calculate: $$\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{  2\sqrt{x}+4\sqrt[8]{x}}}$$
In the numerator, by factoring out x I was able to get to: $$\sqrt{x}\sqrt{1 + \frac{1}{\sqrt{x}}}\sqrt[4]{1 + \frac{1}{\sqrt{x}}}$$
In the denumerator my success is:
$$\sqrt{x}(2 + 4\sqrt[-\frac{3}{8}]{x})$$
So I could cancel out $$\frac{\sqrt{x}}{\sqrt{x}}$$
But it will leave me with 
$$\lim\limits_{x\to 0}{\frac{\sqrt{1 + \frac{1}{\sqrt{x}}}\sqrt[4]{1 + \frac{1}{\sqrt{x}}}}{2 + 4\sqrt[-\frac{3}{8}]{x}}}$$
What to do next?
 [/INST] Note that the function is not real for negative $x$.
$\sqrt[\frac{-3}{8}]{x}=\frac{1}{\sqrt[\frac{3}{8}]{x}}\to \infty$ as $x\to0^+$. So, it is not the way to do.
\begin{align*}
\lim\limits_{x\to 0^+}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{  2\sqrt{x}+4\sqrt[8]{x}}}&=\lim\limits_{x\to 0^+}{\frac{ \sqrt{x+\sqrt[4]{x}\sqrt{\sqrt{x}+1}} }{  2\sqrt{x}+4\sqrt[8]{x}}}\\
&=\lim\limits_{x\to 0^+}{\frac{ \sqrt[8]{x}\sqrt{\sqrt[4]{x^3}+\sqrt{\sqrt{x}+1}} }{  2\sqrt[8]{x}(\sqrt[8]{x^3}+2)}}\\
&=\lim\limits_{x\to 0^+}{\frac{ \sqrt{\sqrt[4]{x^3}+\sqrt{\sqrt{x}+1}} }{  2(\sqrt[8]{x^3}+2)}}\\
&=\frac{1}{4}
\end{align*}
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