Calculate: $\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}$

Question

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?

Answer 1

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*}

Answer 2

Being lazy (to me, this is not a sin !), let $x=y^8$ with $y >0$. Then, after simplifications, $$\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}=\lim\limits_{y\to 0}\frac{\sqrt{y^6+\sqrt{1+y^4}}}{4+2 y^3}\to \frac 14$$