# for $\epsilon\gt0$ find $\delta_\varepsilon\gt0$ such that $\left|x-x_0\right|\lt\delta_\varepsilon\Rightarrow \left|f(x)-f(x_0)\right|<\varepsilon$

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

1. How do we get from step 1 to step 2?
2. What is the idea of approximating $|x|$ and $|x_0|$ like in 3 and 4?
3. How do we get from 2 to 5 by using those approximations?
4. How do we get from 5 to 6?
5. 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.

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The $=$ sign in (2) should be $\leqslant$ and a square should be added in (6). More importantly, I fail to understand why the unnatural (3) and (4) should be required: $|x|\leqslant|x_0|+1$ for every $\delta\leqslant1$ is more direct, leading to $\delta=\min\{1,\varepsilon/(1+3|x_0|+3|x_0|^2)\}$. – Did Nov 30 '12 at 13:51

## 1 Answer

1. Using the identity $(A^3-B^3)=(A-B)(A^2+AB+B^2)$ and the fact that $|AB|=|A|\cdot|B|$.

2. The idea is to get the same thing at many places at once.

3. $|x^2|+|x_0x|+|x_0^2|=|x|\cdot |x|+|x|\cdot |x_0|+|x_0|\cdot |x_0|$. Each of the 3 terms is less than $(|x_0+1|+|x_0-1|)^2$, that's where you get the number $3$ in the bound. Then you just have to apply the outer absolute value.

4. The approximation $(3)$ works only if $|x-x_0|\leq 1$. The last inequality of $(5)$ is true exactly if $|x-x_0|$ is les than the large fraction in $(6)$. Since $|x-x_0|<\delta$, we set $\delta$ to the lesser of these two numbers.

5. The idea is that proving that a function is continuous at some point sometimes takes a lot of "low-level" mathematics. To avoid them, we have theorems, like that every polynomial is continuous or that every function that has a derivative is continuous.

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