Let $X_n$ be a sequence of RV so that $G_n:=\sqrt{n} \left(X_n-1\right) \underset{n \to \infty}{\overset{d}{\longrightarrow}} G \sim N(\mu,\sigma^2)$.
I want to show that in this case $\sqrt{n} \left(1-X_n^{-1}\right)=G_n+o_P(1).$
I had planned to use the function $h(x)=1-1/x, \quad x\ne 0,\quad$ on $X_n \textbf{1}_{\{X_n \ne 0\}}$ and do a Taylor expansion in $1$ but don't know how to handle the case $X_n =0$, even if I split $h$ up into $h|_{x<0}$ and $h|_{x>0}$. From the above follows that $X_n {\overset{P}{\longrightarrow}} 1 $, so $P(X_n =0) \overset{n}{\longrightarrow} 0$ but I'm not sure how to make use of that.
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Here is what I was planning to do so you've got a better idea:
In the case $X_n>0$ I get $1-X_n^{-1}=h(X_n)=h(1)+h'(\xi_n)(X_n-1)=h'(\xi_n)(X_n-1)$ for some $\xi_n$ between $1$ and $X_n$.
From $X_n {\overset{P}{\longrightarrow}} 1$ follows $\xi_n {\overset{P}{\longrightarrow}} 1$ in case $X_n>0$ (is this even true?) and with the Continuous Mapping Theorem I get $h'(\xi_n) {\overset{P}{\longrightarrow}} h'(1)=1,$ so $h(\xi_n)=1+o_P(1).$
Now I have something like
$\sqrt{n} \left( 1-X_n^{-1}\right)=\sqrt{n}\, h(X_n)\textbf{1}_{\{X_n > 0\}}+\sqrt{n} \left( 1-X_n^{-1}\right)\textbf{1}_{\{X_n \le 0\}}$ $=h'(\xi_n)\, G_n\textbf{1}_{\{X_n > 0\}}+\sqrt{n} \left( 1-X_n^{-1}\right)\textbf{1}_{\{X_n \le 0\}}$ $=(1+o_P(1))\, G_n\textbf{1}_{\{X_n > 0\}}+\sqrt{n} \left( 1-X_n^{-1}\right)\textbf{1}_{\{X_n \le 0\}}$
Now I don't know what to do with the rest and I'm not even sure if the Taylor expansion is the right path to take here.
This is a proof for $\mu = 0$. When $\mu \neq 0$, the proof can be done with some slight modification.
You want to prove $\sqrt{n}(1-X_n^{-1}) - G_n = \sqrt{n}(2-X_n^{-1}-X_n)$ converges to $0$ in probability, which is equivalently that it converges to $0$ in distribution.
To do this, we can apply the second order delta method(for reference see this) with the function $g(x) = x + \frac{1}{x}$.
Since $g'(x) = 1 -\dfrac{1}{x^2}$ and $g''(x) = \dfrac{2}{x^3}$, we have $g'(1) = 0$ and $g''(1) = 2$, the second order delta method says:
$$n(g(X_n) - g(1)) \to \sigma^2\chi_1^2 \text{ in distribution}$$
then we have
$$\sqrt{n}(g(X_n) - g(1)) \to 0 \text{ in distribution}$$ which is the desired result.