Assuming that \(x_{n}\to x\) as \(n \to \infty\) where \(x_{n}\neq 0~\forall n\) and \(x\neq 0\). Then: 1. If \(1<\beta\) there exists \(N \in \mathbb{N}\) such that if \(n>N\implies|x_{n}|\geq \frac{|x|}{\beta}\); 2. \(\lim_{ n \to \infty }\left( \frac{1}{x_{n}} \right)= \frac{1}{x}\).
Proof: For (1) we let \(\epsilon = \frac{(\beta-1)|x|}{\beta}>0\) and since \(x_{n}\to x\) as \(n \to \infty\) we have that \[ \exists N \in \mathbb{N} : \forall n>N,~|x_{n}-x|< \frac{(\beta-1)|x|}{\beta}. \] Since, \(|x|-|x_{n}|\leq |x_{n}-x|\) we can rearrange the full expression to give \[ |x|- \frac{{(\beta-1)|x|}}{\beta}<|x_{n}|\implies \frac{|x|}{\beta}<|x_{n}|. \] For (2) take \(\epsilon>0\) and by part (a), for \(\beta=2\) \[ \begin{align} \exists N_{1}\in \mathbb{N}&:\forall n>N_{1},~|x_{n}|> \frac{|x|}{2}\implies \frac{1}{|x_{n}|}< \frac{2}{|x|} \\ \exists N_{2}\in N_{2}&: \forall n>N_{2}, |x_{n}-x|< \frac{|x|^2\epsilon}{2}. \end{align} \] We let \(N=\max(N_{1}, N_{2})\) and considering \(n>N\) we have that \[ |x_{n}-x|< \frac{|x|^2\epsilon}{2}\quad\&\quad \frac{1}{|x_{n}|}< \frac{2}{|x|} , \] thus \[ \left| \frac{1}{x_{n}}- \frac{1}{x}\right|= \left| \frac{{x-x_{n}}}{x x_{n}}\right|= \frac{|x_{n}-x|}{|x| |x_{n}|}<\frac{2|x_{n}-x|}{|x|^2} < \frac{{2|x|^2 \epsilon }}{2} \frac{1}{|x|^2}=\epsilon . \]