Limit proof in real analysis Suppose that $x_n$ is a sequence of real numbers that converges to $1$ as $n\to\infty$. Using definition 2.1, prove that the following limit exists.
Definition 2.1:
"A sequence of real numbers $\{x_n\}$ is said to converge to a real number $a\in\mathbb{R}$ if and only if for every $\epsilon>0$ there is a $N\in\mathbb{N}$ (which in general depends on $\epsilon$) such that $$n\geq N\hspace{.5cm} \text{implies} \hspace{.5cm} |x_n-a|<\epsilon.$$
The sequence is the following: $$\frac{\pi x_n-2}{x_n}\to\pi-2 \hspace{.5cm} \text{as} \hspace{.5cm} n\to\infty.$$
My work:
By hypothesis, given $\epsilon>0$ there is an $N\in\mathbb{N}$ such that $n\geq N$ implies $x_n>\frac{1}{2}$ and $|x_n-1|<\frac{\epsilon}{4}$. In particular $\frac{1}{x_n}<2$. Thus $n\geq N$ implies $$\left|\frac{\pi x_n-2}{x_n}-(\pi-2)\right|\equiv 2\left|\frac{x_n-1}{x_n}\right|.$$
At this point, I don't know where to take the problem. I only got this far from help from other problems I've done before. I know the answer is supposed look something along the lines of $$\left|\frac{\pi x_n-2}{x_n}-(\pi-2)\right|\equiv 2\left|\frac{x_n-1}{x_n}\right|<4|x_n-1|<\epsilon,$$ but I have no idea how I'm supposed to arrive at that solution. Additionally, setting up the problem confuses me because I don't know why we need to specify that $x_n>\frac{1}{2}$ and why in $|x_n-1|<\frac{\epsilon}{4}$ we divide $\epsilon$ by $4$ and not $2.$ Any help clearing up this confusion would be a big help. 
 A: You just say :
$x_n$ is convergent then $\pi x_n-2 $ converges to $\pi-2$ and $ x_n^{-1} $ converges to 1.
By product it is ok.
Furthermore: with def
Let $\epsilon>0$ inferior to 1  . Let $n_0$ from the def . 
First part
$$ \forall n > n_0 \ , \ |x_n-1|<\epsilon $$
$$\forall n > n_0 \ ,  |\pi x_n-2-(\pi-2)|=|x_n-1|\pi<\epsilon\pi$$Second part
$$\forall n > n_0 \ , \ |x_n-1|<\epsilon $$
$$\forall n > n_0 \ , \ |1-\frac{1}{x_n}|<\frac{\epsilon}{x_{n}} $$
or we have (using absolute inequality meaning)
$$\forall n > n_0 \ , \ x_n>1-\epsilon $$
so 
$$\forall n > n_0 \ , \ x_n^{-1}<\frac{1}{1-\epsilon} $$
hence 
$$\forall n > n_0 \ , \ |1-\frac{1}{x_n}|<\frac{\epsilon}{x_{n}}<\frac{\epsilon} {1-\epsilon}$$
Conclusion
$$\forall n > n_0 \ , |\frac{\pi x_n-2-(\pi-2)}{x_n}|<\epsilon\pi\frac{\epsilon} {1-\epsilon}$$
Conclude by setting $ \epsilon'= \epsilon\pi\frac{\epsilon}{1-\epsilon} $
A: Let $\epsilon > 0$. Since $(x_n)_{n=1}^{\infty}$ converges to $1$, there exists $N_1 \geq 1$ so that $x_n > \frac{1}{2}$ for all $n > N_1$. This helps you bound the denominator of your expression. It is also true that there exists $N_2 \geq 1$ so that $|x_n - 1| < \frac{\epsilon}{4}$ for $n > N_2$. Let $N = \max\{N_1, N_2\}$. Then for $n > N$, we have $$ \left| \frac{\pi x_n - 2}{x_n} - (\pi - 2) \right| = \left| \frac{\pi x_n - 2}{x_n} - \frac{x_n(\pi - 2)}{x_n} \right| = \left| \frac{2x_n - 2}{x_n} \right| = 2 \left| \frac{x_n - 1}{x_n} \right|,$$ and $$ 2 \left| \frac{x_n - 1}{x_n} \right| < 2 \left| \frac{x_n - 1}{\frac{1}{2}} \right| = 4 |x_n - 1| < 4 \cdot \frac{\epsilon}{4} = \epsilon.$$
You use $\frac{\epsilon}{4}$ because the final term is multiplied by 4. 
