- Prove that
![\[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)) \quad \text{as} \quad x \to 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fd8baf4cf63dab00a9fc7747b181a36e_l3.png "Rendered by QuickLaTeX.com")
where
as 
.
- Using part (a) prove
![\[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + o(x^5) \quad \text{as} \quad x \to 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3871cb8329f483ea952ffebd70a2199c_l3.png "Rendered by QuickLaTeX.com")
- Proof. We start with the algebraic identity
![\[ \frac{1}{1+u} = 1 - u + u^2 - \frac{u^3}{1+u}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9071562b1476e36b9822aaad56ad155d_l3.png "Rendered by QuickLaTeX.com")
Replacing
by 
we have,![\[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) - \frac{g^3(x)}{1+g(x)} = 1 - g(x) + g^2(x) - g^2(x) \frac{g(x)}{1+g(x)}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fd93992fd16d51de93b739498e96fdca_l3.png "Rendered by QuickLaTeX.com")
By the definition of
we have ![\[ g(x) = o(1) \text{ as } x \to 0 \quad \implies \quad \lim_{x \to 0} g(x) = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0b6b542c231063d285d17120ba4f05ae_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\[ \lim_{x \to 0} \frac{g(x)}{1+g(x)} = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4a1f87bbc878dc88122c8955a0d00cee_l3.png "Rendered by QuickLaTeX.com")
Hence,
![\[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)). \qquad \blacksquare\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-af74f7c0bcbef9b625b4b34f42ea41af_l3.png "Rendered by QuickLaTeX.com")
- Proof. First, we use the expansion of
,
![\[ \frac{1}{\cos x} = \frac{1}{1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)} = \frac{1}{1+g(x)} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e7f567ea3f458496cc482b60cb7bf0c7_l3.png "Rendered by QuickLaTeX.com")
where
. By part (a) we then have as ,![\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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d90a756614312d9d5202b27a464afb0a_l3.png "Rendered by QuickLaTeX.com")
Next, since
we multiply this expansion for 
by the expansion (page 287 of Apostol) for 
as ,![\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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8d91ac2660a0df865c50c181d34fe3b2_l3.png "Rendered by QuickLaTeX.com")
I think there is a mistake in the 1/cos(x) line. You didn’t write o(x^5) at the end and even if you did, when expanding you get x^4/4 as the lowest power term and it’s not little oh of x^4, so it reduces to o(x^3) which is gonna ruin the rest of the answer.