Suppose that $\{x_n\}$ is a decreasing sequence having limit $0$. How to find the reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ as $n\to\infty$.
In my textbook it says that \begin{equation}\frac{1}{x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))}=\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1) \;(n\to\infty). \end{equation}
That is,
\begin{align*}
&[x_{n-1}^2-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)]\cdot[\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1)] \\
=&1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2)+\frac{x_{n-1}^2}{3}-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)+o(x_{n-1})+o(x_{n-1}^3)+o(x_{n-1}^4) \\
=&1+o(x_{n-1})\;(n\to0).
\end{align*}
Also,I find that
\begin{align*}
&[x_{n-1}^2-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)]\cdot\frac{1}{x_{n-1}^2} \\
=&1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2) \\
=&1+o(x_{n-1})\;(n\to 0).&
\end{align*}
So my question is that why the reciprocal should be $\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1)$.
This comes out from the geometric series ($x\rightarrow 0$) $$ \frac{1}{1-x}=1+x+O(x^2). $$ In your case, you have to evaluate $$ \frac{1}{x_n^2\left(1-\frac{x^2_n}{3}+O(x^4_n)\right)} $$ Using the geometric series you will get $$ \frac{1}{x_n^2}\left(1+\frac{x_n^2}{3}+O(x_n^4)\right)=\frac{1}{x_n^2}+\frac{1}{3}+O(x_n^2) $$ that is the result in the textbook.