Doubt in proof of continuity I was reading some examples of proving continuity using $\epsilon$-$\delta$ argument and well, I've found one that I'm not understanding one step. The problem is: prove that the function $f: \mathbb{R}-\{0\} \to \mathbb{R}$ given by:
$$f(x)=\frac{1}{x}-\frac{1}{x_0}$$
Is continuous for every $x_0 \in \mathbb{R} - \{0\}$. Well, the given solution is as follows. First we note that we have:
$$f(x)-f(x_0)=\frac{x_0-x}{x_0x}$$
Now here comes the problem, it says: if $x_0 > 0$, then for every $x\in I=(x_0/2, 3x_0/2$) we have:
$$x x_0>\frac{x_0}{2} x_0 = \frac{x^2_0}{2} \quad \Longrightarrow \quad \left|\frac{1}{x}-\frac{1}{x_0}\right| = \frac{\left|x_0-x\right|}{{x_0x}}<2\frac{\left|x_0-x\right|}{x_0^2}$$
And then this is used on the proof. That's fine, but where did this interval come from? I understood that if $x \in I$ then obviously $x > x_0/2$ which justifies the line below, but my problem is, where did the interval come from? I know that for every point on the real line we can surround it with an open interval, but well, this is that kind of trick that I look and I think: "oh my, I would never think about it.", so what's the reasoning behind this?
Thanks very much in advance for the help.
 A: you don't need to use the specified interval, you can think about continuity as follows if the points are "close" to each other then their  images must be "close" to each other.let $\epsilon >0$ be given want to show that if $x$ and $x_0>0$ are close to each other then their images are close, we have $$|f(x)-f(x_0)|=\frac{|x-x_0|}{|x||x_0|}$$ let $\delta<\epsilon$ and $|x-x_0|<\delta $ , the only thing left is to find a lower bound for $|x|$ ,we have $-\delta+x_0<x<\delta+x_0$ which implies that $\frac{1}{x}<\frac{1}{x_0-\delta}$( for this to make sense we must assume that $x_0>\delta$), lets recap, let $\epsilon>0$ be given choose  $\delta <min\{\epsilon,x_0\}$ then for $|x-x_0|<\delta$ (i.e the points are close to each other) we have$$|f(x)-f(x_0)|=\frac{|x-x_0|}{|x||x_0|}<\frac{\epsilon}{|(x_0-\delta)||x_0|}<\frac{\epsilon}{|x_0|^2}$$i.e the images are close to each other. which implies that $f$ is continuous at $x_0$. you can do the same thing for $x_0<0$ which completes the proof.
A: The specific interval isn't really important.  You just need an interval that contains $x_0$ and doesn't contain zero.  The argument you write would need to be modified a bit, but the same idea would go through.  As vadim123 points out, the "author" of the argument just took some natural interval with these properties.
