<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] The expression under limit can be written as $$\frac{y^{1/3}-1}{x}=\frac{y^{1/3}-1}{y-1}\cdot\frac{y-1}{x}\tag{1}$$ where $$y=\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}=\frac{x}{1+\sqrt{1-x}}\cdot\frac{\sqrt{1+x}+1}{x}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}\tag{2}$$ Since $y\to 1$ as $x\to 0$ we can see that the first fraction in $(1)$ tends to $1/3 $ via the standard formula $$\lim_{t\to a} \frac{t^n-a^n}{t-a}=na^{n-1}\tag{3}$$ Hence the desired limit is equal to the limit of $$\frac{1}{3}\cdot\frac{y-1}{x}=\frac{\sqrt {1+x}-\sqrt{1-x}}{3x(1+\sqrt{1-x})}=\frac{2x}{3x}\cdot\frac{1}{(\sqrt{1+x}+\sqrt{1-x})(1+\sqrt{1-x})}$$ It follows that the desired limit is $$\frac{2}{3}\cdot\frac{1}{2\cdot 2}=\frac{1}{6}$$ In general avoid rationalization of expressions containing radicals beyond square roots and instead use the limit formula $(3)$. This saves a lot of typing effort and reduces visual clutter.
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