$x_n$ is a sequence of real numbers that converges to 1 as $n\to\infty$
How to prove$$\lim_\limits{n\to\infty}\left(\frac{\pi x_n -2}{ x_n}\right)=\pi -2$$ by the formal definition of sequence limits
Since $\forall \epsilon\gt 0\; \exists N \in\mathbb{N}$ such that$\forall n\geqslant N\Rightarrow|x_n -1|<\epsilon$
I've tried$|\left(\frac{\pi x_n -2}{ x_n}\right)-(\pi -2)| = 2|\frac{x_n -1}{ x_n}|<\epsilon_1$
$|x_n -1|<max(1,x_n)\frac{\epsilon_1}{2}\Rightarrow -max(1,x_n)\frac{\epsilon_1}{2}+1<x_n$
Let$k(x_n)=n$
consider $N_1=\lceil k(-max(1,x_n)\frac{\epsilon_1}{2}+1)\rceil$ then$\lim_\limits{n\to\infty}\left(\frac{\pi x_n -2}{ x_n}\right)=\pi -2$
Is the procedure correct?
Notice that $\frac{\pi - 2x_{n}}{x_n}$ = $\frac{\pi}{x_n} - 2$. So it suffices to show that given a sequence ${s_n}$ with $s_n \neq 0 \forall n \in \mathbb{N}$ and $s_n \to s$, $s \neq 0$, then $\frac{1}{s_n} \to \frac{1}{s}$.
Choose $N_1 \in \mathbb{N}$ such that $\forall n \geq N_1$, $|s_n - s| < \frac{1}{2}|s|$ $\implies |s_n|> \frac{1}{2}|s|$ (by triangle inequality).
Choose $N_2 \in \mathbb{N}$ such that $\forall n \geq N_2$, $|s_n -s| < \frac{1}{2}|s|^2\epsilon$.
Take $N = \max\{N_1,N_2\}$. $\forall n \geq N$, it follows that
$|\frac{1}{s_n} - \frac{1}{s}| = \frac{|s-s_n|}{ss_n} < \frac{2}{|s|^2}|s-s_n| < \epsilon$ .
So $\lim_\limits{n\to\infty} \frac{\pi - 2x_{n}}{x_n} = \lim_\limits{n\to\infty} \frac{\pi}{x_n} - 2 = \pi\lim_\limits{n\to\infty}\frac{1}{x_n} - 2 = \pi - 2$.