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...
 A: Using:
$$
   1 -\sqrt{1-x} = \frac{x}{1+\sqrt{1-x}}\qquad \frac{1}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{x}
$$
we have:
$$
  \frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{1+\sqrt{1-x}} = 1 + \frac{\sqrt{1+x}-\sqrt{1-x}}{1+\sqrt{1-x}} = 1 + \underbrace{\frac{2x}{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(1+\sqrt{1-x}\right)}}_{w}
$$
And using
$$
    \sqrt[3]{1+w} - 1 = \frac{w}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} }
$$
we have
$$
  \lim_{x \to 0}\frac{1}{x} \left(\sqrt[3]{ \frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} } - 1\right) = \lim_{x \to 0} \frac{w}{x} \cdot \frac{1}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} } = \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{6}
$$
A: 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:
$$\lim_{x\rightarrow 0} \frac{1}{x}\left((1+x/2+\mathcal{O}(x^2))^{1/3}-1\right)=\lim_{x\rightarrow 0}\frac{1}{x}\left(x/6+\mathcal{O}(x^2)\right)=\frac{1}{6}$$
A: 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.
A: Let us start by noting that:

*

*$\sqrt[3]{1-\sqrt{1-x}}\cdot \sqrt[3]{1+\sqrt{1-x}}=\sqrt[3]{x}$

*$\sqrt[3]{\sqrt{1+x}-1}\cdot \sqrt[3]{\sqrt{1+x}+1}=\sqrt[3]{x}$

*$$\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}=1$$
Using these algebraic facts, we can write that: $$\begin{align} \lim_{x\to 0} \frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\Biggl]
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{(1-\sqrt{1-x})(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\Biggl[\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x}\Biggl] \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{x}}\Biggl] \\  
&= \lim_{x\to 0}\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x} \cdot \lim_{x\to 0} \sqrt[3]{\frac{1-\sqrt{1-x}}{x}} \\ \end{align}$$
If we set $-\sqrt{1-x}$ and $\sqrt[3]{\sqrt{1+x}+1}$ to be $f(x)$ and $g(x)$ respectively, it can be written that: $$\begin{align} \lim_{x\to 0} \frac{g(x)-g(-x)}{x} \cdot\lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \tag{1}\label{eq1}
&= \lim_{x\to 0} 2\Biggl[\frac{g(x)-g(-x)}{2x}\Biggl]\cdot \lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \\
&= 2g'(0)\cdot \sqrt[3]{f'(0)} \tag{2}\label{eq2} \\
&= \Biggl[\frac{(\sqrt{1+x}+1)^{-\frac{2}{3}}}{3\sqrt{1+x}}\Biggl]_{x=0}\cdot \Biggl[\sqrt[3]{\frac{1}{2\sqrt{1-x}}}\Biggl]_{x=0} \tag{3}\label{eq3} \\
&= \frac{1}{3\sqrt[3]{4}}\cdot \frac{1}{\sqrt[3]{2}} \\
&= \frac{1}{6} \\   \end{align}$$
$(2)$: limit definition of the derivative
Hope it helps!
