<s>[INST] Prove that $ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$ 
Prove that $$ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$$ 

Let $x=y^8$ with $t>0$. Then after we get 
$$
\lim \frac{ \sqrt { t^8 + \sqrt{ t^8 + \sqrt{t^8}}}}{ t} = \lim \frac{ \sqrt{ t^8 + \sqrt{ t^8 + t^4}}}{ t}
$$
  and can't go further than that
 [/INST] We have that
$$ \frac{\sqrt{x + \sqrt{ x + \sqrt x}}}{\sqrt[8] x} = \frac{ \sqrt{x + \sqrt[4] x\sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} = \frac{\sqrt[8]x\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} =$$
$$=\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}} \to \sqrt{0 + \sqrt{ 0+1}}=1$$
or in a similar way by your approach
$$\frac{ \sqrt{ t^8 + \sqrt{ t^8 + t^4}}}{ t}=\frac{ \sqrt{ t^8 + t^2\sqrt{ t^4 + 1}}}{ t}=\frac{ t\sqrt{ t^6 + \sqrt{ t^4 + 1}}}{ t}=\sqrt{ t^6 + \sqrt{ t^4 + 1}} \to 1$$
</s>