For the following function, for each $\varepsilon\gt0$ find $\delta_\varepsilon\gt0$ such that $\left|x-x_0\right|\lt\delta_\varepsilon\Rightarrow \left|f(x)-f(x_0)\right|<\varepsilon$, where $D(f)=\mathbb{R}$ $f(x)=x^3$
The solution looks like this
$\left|f(x)-f(x_0)\right| = \left|x^3-x_0^3\right| = \tag{1}$
$=\left|x-x_0\right|\cdot \left| \left|x\right|^2 + \left|x\right| \cdot \left|x_0\right|+ \left|x_0\right|^2 \right| \tag{2}$
We approximate:
$ \left|x\right| \le \left|x_0+1\right| + \left|x_0-1\right| \tag{3} $
$ \left|x_0\right| \le \left|x_0+1\right| + \left|x_0-1\right| \tag{4} $
so together with relation $(2)$ we get
$ \left|x-x_0\right|\cdot \left| \left|x\right|^2 + \left|x\right| \cdot \left|x_0\right|+ \left|x_0\right|^2 \right| \le \left|x-x_0\right| \cdot 3 \cdot (\left|x_0-1\right| + \left|x_0 + 1\right|)^2 \le \varepsilon \tag{5} $
So we get
$ \delta = \min\left\{ 1, \frac{\varepsilon}{3\left(\left|x_0-1\right|+\left|x_0+1\right|\right)} \right\} \tag{6} $
Questions
- How do we get from step 1 to step 2?
- What is the idea of approximating $|x|$ and $|x_0|$ like in 3 and 4?
- How do we get from 2 to 5 by using those approximations?
- How do we get from 5 to 6?
- What is the whole "idea behind the maths" done here? Like the overall story line?
It feels like it's at the heart of limits but I kind of miss important points. Please help me "connect the dots" in a conceptual way.
I would like to gain some insight, not just memorize it.