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$x_n$ 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?

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    $\begingroup$ As $(\pi x-2)/x=\pi-2/x$ it is enough to deal with $2/x$. $\endgroup$ – Yves Daoust Oct 13 '18 at 8:12
  • $\begingroup$ Can you use that $\varepsilon-\delta$ definition of continuity is equivalent to the definition with limits? Because this asks you precisely to prove that your function is continuous at $1$. $\endgroup$ – Ennar Oct 13 '18 at 9:47
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Notice that $\frac{\pi - 2x_{n}}{x_n}$ = $\pi - \frac{2}{x_n}$. 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 $\lim_\limits{n\to\infty} \frac{\pi - 2x_{n}}{x_n} = \lim_\limits{n\to\infty} \pi - \frac{2}{x_n} = \pi - 2\lim_\limits{n\to\infty}\frac{1}{x_n} = \pi - 2$.

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