Reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ where $x_n\searrow0$ 
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.
