Prove that 1 / (1+g(x)) = 1- g(x) + g2(x) + o(g2(x))

Prove that
$\frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)) \quad \text{as} \quad x \to 0$

where $g(x) = o(1)$ as $x \to 0$ .
Using part (a) prove
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + o(x^5) \quad \text{as} \quad x \to 0.$

Proof. We start with the algebraic identity
$\frac{1}{1+u} = 1 - u + u^2 - \frac{u^3}{1+u}.$

Replacing $u$ by $g(x)$ we have,

$\frac{1}{1+g(x)} = 1 - g(x) + g^2(x) - \frac{g^3(x)}{1+g(x)} = 1 - g(x) + g^2(x) - g^2(x) \frac{g(x)}{1+g(x)}.$

By the definition of $o(1)$ we have

$g(x) = o(1) \text{ as } x \to 0 \quad \implies \quad \lim_{x \to 0} g(x) = 0.$

Therefore,

$\lim_{x \to 0} \frac{g(x)}{1+g(x)} = 0.$

Hence,

$\frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)). \qquad \blacksquare$
Proof. First, we use the expansion of $\cos x$ ,
$\frac{1}{\cos x} = \frac{1}{1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)} = \frac{1}{1+g(x)}$

where $g(x) = -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5)$ . By part (a) we then have as $x \to 0$ ,

$\begin{align*} \frac{1}{\cos x} &= 1 - g(x) + g^2(x) + o(g^2(x)) \\[9pt] &= 1 - \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right) + \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right)^2 + o \left( \left( -\frac{x^2}{2} + \frac{x^4}{24} \right)^2 \right) \\[9pt] &= 1 + \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^4}{4} + o(x^4) \\[9pt] &= 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4). \end{align*}$

Next, since $\tan x = \frac{\sin x}{\cos x}$ we multiply this expansion for $\frac{1}{\cos x}$ by the expansion (page 287 of Apostol) for $\sin x$ as $x \to 0$ ,

$\begin{align*} \tan x &= \frac{1}{\cos x} \cdot \sin x \\[9pt] &= \left( 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4) \right) \left( x - \frac{x^3}{6} + \frac{x^5}{120} + o(x^6) \right) \\[9pt] &= x - \frac{x^3}{3} + \frac{x^5}{120} + \frac{x^3}{2} - \frac{x^5}{12} + \frac{5x^5}{24} + o(x^5) \\[9pt] &= x + \frac{x^3}{6} + \frac{2x^5}{15} + o(x^5). \qquad \blacksquare \end{align*}$