Why can you prove the continuity of $f(x) = \frac{x}{1+x²}$ with $\delta = \min \{ 2, \frac{\epsilon}{2} \} $? Let's take my question as an example. I just don't get it.
What does $\delta = \min \{ 2, \frac{\epsilon}{2} \} $ mean ? (especially 'min{}')
 A: It turns out $\delta=\varepsilon/2$ is plenty in this particular case, so you don't need this sort of argument here.
A better example is to show that $f(x)=x^2$ is continuous at $1$. We do some algebra:
$$|f(1)-f(y)|=|1-y^2|=|1+y||1-y|.$$
Since this has a factor of $|1-y|$, the naive thing to do is to make $\delta$ depend linearly on $\varepsilon$. This has a problem: when $\varepsilon$ is large, $\delta$ will be too large, because $|1+y|$ will grow. For example, taking $\delta=\varepsilon/3$ will not work for $\varepsilon=300$, because for $y=101$ we get $|1-y^2|$ being something like $10000$.
So we need to keep $|1+y|$ moderately small, which we can do only by controlling $|1-y|$. One way is to require $|1-y|<1$ regardless of $\varepsilon$. This works because $|1-y|<1$ is equivalent to $0<y<2$, so we guarantee $1<1+y<3$ and in particular $|1+y|<3$.
This means that if we require $|1-y|<1$ then we have $|1-y^2|<3|1-y|$. We want this to be less than our given $\varepsilon$, so we further require $|1-y|<\varepsilon/3$ to get $|1-y^2|<3 \varepsilon/3 = \varepsilon$ as desired.
To get $|1-y|<1$ and $|1-y|<\varepsilon/3$, we take $\delta=\min \{ 1,\varepsilon/3 \}$.
A: 
What does $\delta = \min \{ 2, \frac{\epsilon}{2} \} $ mean ? (especially 'min{}')

$$\min\left\{ 2, \frac{\epsilon}{2}\right\}=\left\{\begin{array}{ccc}2&\text{if}&\epsilon\gt4\\\frac{\epsilon}{2}&\text{if}&\epsilon\leqslant4\end{array}\right.$$
A: Let $\epsilon>0$ be given. Choose $δ:=\min\{\frac\epsilon2,2\}$. Then:
$$\left|f(x)-f(x_0)\right|=\left|\frac{x}{1+x^2}-\frac{x_0}{1+{x_0}^2}\right|=$$
$$=\left|\frac{(xx_0-1)(x_0-x)}{(1+x^2)(1+{x_0}^2)}\right|<\delta\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|$$
Now let's show that for any $\delta: \space \left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2$
( I'll show only $\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}<2$, it's not hard to show that it's $>-2$ )
$$\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}<\frac{(x_0+\delta)x_0-1}{(1+(x_0-\delta)^2)(1+{x_0}^2)}$$
Let's assume:
$$\frac{(x_0+\delta)x_0-1}{(1+(x_0-\delta)^2)(1+{x_0}^2)}>2$$
Then:
$$\frac{\delta^2(2x_0^2+2)-\delta(4x_0^3+5x_0)+(2x_0^4+3x_0^2+3)}{(1+(x_0-\delta)^2)(1+{x_0}^2)}<0$$
The denominator and numerator are always positive. Contradiction.
So $\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2$. Then:
$$\left|f(x)-f(x_0)\right|<\delta\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2\delta\leq\epsilon$$
Then if for $\epsilon>2$ we pick $\delta = 2$ and $\delta = \frac\epsilon2$ otherways the inequality holds.
