How to calculate $\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$

Question

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? $$\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$$

Answer 1

Using the binomial expansion, we have

$$\sqrt{1+x}=1+\frac12 x-\frac18x^2+O(x^3)$$

Therefore,

$$g(x)=\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}=\frac{\frac12 x+\frac18x^2+O(x^3)}{\frac12x-\frac18x^2+O(x^3)}=1+\frac12x+O(x^2)$$

Now, using the binomial expansion again reveals that

$$g^{1/3}(x)-1=\frac16x+O(x^2)$$

Dividing by $x$ and letting $x\to 0$ yields the result $\frac16$.

Answer 2

Here's a more elementary approach. Since \begin{eqnarray} \frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}&=& \frac{(1-\sqrt{1-x})\color{blue}{(1+\sqrt{1-x})}\color{green}{(\sqrt{1+x}+1)}}{(\sqrt{1+x}-1)\color{green}{(\sqrt{1+x}+1)}\color{blue}{(1+\sqrt{1-x})}}=\frac{(1-1+x)(\sqrt{1+x}+1)}{(1+x-1)(1+\sqrt{1-x})}\\ &=&\frac{x(\sqrt{1+x}+1)}{x(1+\sqrt{1-x})}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}, \end{eqnarray} using the identity $$ \sqrt[3]{a}-1=\frac{a-1}{\sqrt[3]{a^2}+\sqrt[3]{a}+1}, $$ with $$ a=a(x)=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}} $$ we have: \begin{eqnarray} \frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)&=&\frac{1}{x}\left(\sqrt[3]{\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}}-1\right)=\frac{a-1}{x(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\ &=&\frac{\sqrt{1+x}-\sqrt{1-x}}{x(1+\sqrt{1-x})(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\ &=&\frac{(\sqrt{1+x}-\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\ &=&\frac{1+x-(1-x)}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\ &=&\frac{2x}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\ &=&\frac{2}{(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}. \end{eqnarray} Taking the limit and using the fact that $$ \lim_{x\to 0}a(x)=1, $$ we get: $$ \lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)=\frac{2}{2\cdot2\cdot3}=\frac16. $$

Answer 3

As Mercy notes,

$${1-\sqrt{1-x}\over\sqrt{1+x}-1}={\sqrt{1+x}+1\over1+\sqrt{1-x}}$$

(which can be verified by cross multiplying: $(1-\sqrt{1-x})(1+\sqrt{1-x})$ and $(\sqrt{1+x}-1)(\sqrt{1+x}+1)$ both simplify to $x$). Let

$$u(x)={\sqrt{1+x}+1\over1+\sqrt{1-x}}$$

It's clear that $u(0)=1$, and it's straightforward (if a bit tedious) to calculate $u'(0)={1\over2}$. Consequently, L'Hopital gives us

$$\lim_{x\to0}{u(x)^{1/3}-1\over x}={{1\over3}u(0)^{-2/3}u'(0)\over1}={1\over6}$$

Added later: For the sake of completeness, here's an expedited calculation of $u'(0)$:

Let $f(x)=\sqrt{1+x}+1$ and $g(x)=1+\sqrt{1-x}$. Since $f(0)=g(0)=2$, we have

$$u'(0)={2f'(0)-2g'(0)\over4}={f'(0)-g'(0)\over2}$$

so it suffices to take the derivatives $f'(x)={1\over2}(1+x)^{-1/2}$ and $g'(x)=-{1\over2}(1-x)^{-1/2}$ and note that $f'(0)-g'(0)={1\over2}-(-{1\over2})=1$. (The point is, since all we need is its value at $0$, there's no need to grind out a full, formal expression for $u'(x)$.)

Answer 4

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 $x \to 0$.