# proof$\lim_\limits{n\to\infty}\left(\frac{\pi x_n -2}{ x_n}\right)=\pi -2$

$$x_n$$ 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$$|\left(\frac{\pi x_n -2}{ x_n}\right)-(\pi -2)| = 2|\frac{x_n -1}{ x_n}|<\epsilon_1$$

$$|x_n -1|
Let$$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?

• As $(\pi x-2)/x=\pi-2/x$ it is enough to deal with $2/x$. – Yves Daoust Oct 13 '18 at 8:12
• 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 $1$. – Ennar Oct 13 '18 at 9:47

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 $$\lim_\limits{n\to\infty} \frac{\pi - 2x_{n}}{x_n} = \lim_\limits{n\to\infty} \pi - \frac{2}{x_n} = \pi - 2\lim_\limits{n\to\infty}\frac{1}{x_n} = \pi - 2$$.