Choosing delta in a simple $\delta$-$\varepsilon$ proof.

Question

I need to prove that $f(x)=\begin{cases}|x|, & \text{when } x \leq 0\\ x^2, & \text{when } x > 0 \end{cases}$ is continuous with $\varepsilon$-$\delta$.

I have a solution for this but I don't understand why we should choose the delta in the following way.

When we assume that $x_0 <0$ and then choose $\delta = \min\{-x_0,\varepsilon\}$ we get $|f(x)-f(x_0)|=|x-x_0|<\varepsilon$, when $|x-x_0|< \delta$.

Why don't we just choose $\delta = \varepsilon$? Where do we need the $-x_0$?

Same is done when we consider $x_0 > 0$ and we choose $\delta = \min \{1, x_0, \varepsilon / (2|x_0|+1)\}$. And again same question: Why we need to take $x_0$ here in the minimum?

Answer 1

Hint: What is $f(x)$ when $|x-x_0|>-x_0$?

Answer 2

If $\delta>-x_0$, you have $x_0+\delta>0$. For example, if $x_0=-1$ and $\epsilon=5$, what happens if you choose $\delta=\epsilon?$ Then $x_0+\delta=4$ and that means you want $|f(x)-f(-1)|<3$ when $-6<x<4$. What happens when $x=3$?

The same is true in the other direction, as well - you are basically trying to make sure that $\delta$ is chosen "small enough" so that you don't have to worry about other complications, like values where the "other" function is defined.