<s>[INST] Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...
 [/INST] Multiply $\frac{1-\sqrt{1-x}}{\sqrt{x+1}-1}$ by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ and simplify to get $\frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x}$
Expanding this, you only need keep two terms and you get
$$ \frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x} = \frac{x+x^2/2 + \mathcal{O}(x^3)}{x}$$
then your limit becomes:
$$\lim_{x\rightarrow 0} \frac{1}{x}\left((1+x/2+\mathcal{O}(x^2))^{1/3}-1\right)=\lim_{x\rightarrow 0}\frac{1}{x}\left(x/6+\mathcal{O}(x^2)\right)=\frac{1}{6}$$
</s>