Prove that, as x to 0, cube root of (1+x)-1- frac13x ≈

Prove that, as $x\to0$, $\displaystyle\sqrt[3]{1+x}-1-\frac13x \;\approx\; -\frac{x^{2}}{9}$.

Step 1: State the claim as a limit

Two infinitesimals $\alpha(x)$ and $\beta(x)$ are equivalent as $x\to0$, written $\alpha\approx\beta$, precisely when $\displaystyle\lim_{x\to0}\frac{\alpha(x)}{\beta(x)} = 1$. So it suffices to show

\[ \lim_{x\to0}\frac{\sqrt[3]{1+x}-1-\dfrac{x}{3}}{-\dfrac{x^{2}}{9}} = 1. \tag{1} \]

Step 2: Expand the cube root

The generalised binomial series $\bigl(1+x\bigr)^{\alpha} = \displaystyle\sum_{n\ge0}\binom{\alpha}{n}x^{n}$ with $\alpha = \dfrac13$ gives

\[ \bigl(1+x\bigr)^{1/3} = 1+\frac13x+\frac{\dfrac13\left(\dfrac13-1\right)}{2!}x^{2}+O\!\left(x^{3}\right), \tag{2} \]

and the quadratic coefficient is

\[ \frac{\dfrac13\cdot\left(-\dfrac23\right)}{2} = \frac{-\dfrac29}{2} = -\frac19. \tag{3} \]

Hence \[\bigl(1+x\bigr)^{1/3} = 1+\frac{x}{3}-\frac{x^{2}}{9}+O\!\left(x^{3}\right). \tag{4}\]

Step 3: Isolate the infinitesimal

Subtracting $1+\dfrac{x}{3}$ — that is, subtracting the tangent line at the origin — from both sides of (4): \[\sqrt[3]{1+x}-1-\frac{x}{3} = -\frac{x^{2}}{9}+O\!\left(x^{3}\right). \tag{5}\]

Step 4: Verify the equivalence

\[ \frac{\sqrt[3]{1+x}-1-\dfrac{x}{3}}{-\dfrac{x^{2}}{9}} = \frac{-\dfrac{x^{2}}{9}+O\!\left(x^{3}\right)}{-\dfrac{x^{2}}{9}} = 1+O(x) \longrightarrow 1, \tag{6} \]

which is exactly (1). The equivalence is established.

Step 5: Verification

\[x = 0.1:\ \sqrt[3]{1.1} = 1.0322801,\ \ 1+\tfrac{0.1}{3}-\tfrac{0.01}{9} = 1.0322222 \;\;\checkmark\]

\[\text{Ratio in (1): } x = 0.1:\ 0.9478961; \qquad x = 0.01:\ 0.9944812; \qquad x = -0.1:\ 1.0595539 \;\;\checkmark\]

\[\text{Next term: } \bigl(1+x\bigr)^{1/3} = 1+\frac{x}{3}-\frac{x^{2}}{9}+\frac{5x^{3}}{81}-\cdots \;\;\checkmark\]