How to calculate $\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$ I've been studying limits on Rudin, Principles of Mathematical Analysis for a while, but the author doesn't exactly explain how to calculate limits...so, can you give me a hint on how to solve this? $$\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$$

Using the basic limit $$\lim_{t \to a}\frac{t^{n} - a^{n}}{t - a} = na^{n - 1}\tag{1}$$ we can see by putting $n = 1/2, a = 1, t = 1 + x$ that $$\lim_{x \to 0}\frac{\sqrt{1 + x} - 1}{x} = \frac{1}{2}\tag{2}$$ Replacing $x$ by $-x$ we get $$\frac{1 - \sqrt{1 - x}}{x} = \frac{1}{2}\tag{3}$$ From the equation $(2), (3)$ we get (by division) $$\lim_{x \to 0}\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1} = 1\tag{4}$$ If $$u = \frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}$$ then $u \to 1$ as $x \to 0$ and hence by $(1)$ we get $$\lim_{u \to 1}\frac{\sqrt[3]{u} - 1}{u - 1} = \frac{1}{3}$$ Our desired limit is then given by
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[3]{u} - 1}{x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[3]{u} - 1}{u - 1}\cdot\frac{u - 1}{x}\notag\\
&= \lim_{u \to 1}\frac{\sqrt[3]{u} - 1}{u - 1}\cdot\lim_{x \to 0}\frac{u - 1}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x\{\sqrt{1 + x} - 1\}}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x}{\sqrt{1 + x} - 1}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x^{2}}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{4 - \{\sqrt{1 - x} + \sqrt{1 + x}\}^{2}}{x^{2}\{2 + \sqrt{1 - x} + \sqrt{1 + x}\}}\notag\\
&= \frac{1}{6}\lim_{x \to 0}\frac{4 - \{2 + 2\sqrt{1 - x^{2}}\}}{x^{2}}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \sqrt{1 - x^{2}}}{x^{2}}\notag\\
&= \frac{1}{3}\cdot\frac{1}{2}\text{ (from equation }(3))\notag\\
&= \frac{1}{6}\notag\\
\end{align}
Note: The final limit calculation can be simplified greatly if we note that $u$ can be expressed as $$u = \frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1} = \frac{\sqrt{1 + x} + 1}{\sqrt{1 - x} + 1}$$ and then it is very easy to calculate the limit of $(u - 1)/x$ as $x \to 0$.
