Help needed in the details of proving continuity for the function $f(x)=\frac{1}{x^2}$ on $(0,\infty)$

Question

Here is the example taken from Ross' Elementary Analysis ((*) refers to the $\epsilon$-$\delta$ definition of continuity):

Usually it is possible to know what to value to bound $|x-x_0|$ by (usually a natural number). Where is the intuition to let $|x-x_0|$ be bounded by $\frac{x_0}{2}$? Furthermore in the line that starts with "If $|x-x_0|$...", I'm not sure how $|x-x_0|<\frac{x_0}{2}$ leads to the remainder of the sentence aside from $|x|<|x_0|+\frac{x_0}{2}=|x|<\frac{3x_0}{2}$. Any explanations are appreciated.

Answer 1

You do not want $|x-x_0| \ge x_0$ as that could allow $x \le 0$.

Nor would you want $x$ arbitrarily small and positive, as that would allow $\frac{1}{x^2}-\frac{1}{x_0^2}$ to be arbitrarily big.

You can avoid both of these by restricting $|x-x_0|$ to be some fraction of $x_0$, even before considering $\epsilon$.

$\dfrac{x_0}{2}$ is the simplest such fraction.

Answer 2

If $|x-x_0|<x_0$ then $x \in (0\ ,2x_0)$. If $x$ would be in any interval $(0,a)$ then $\frac{1}{x}$ can be chosen in $(\frac{1}{a},\infty)$ which we can't allow. We need an interval farther from zero.

If $|x-x_0|<\frac{1}{2}x_0$ then $x \in (\frac{1}{2}x_0,\frac{3}{2}x_0)$. This is an open interval between two values that are clearly not zero. This is of course not unique in any way, as any constant $c \in (0\ , 1)$ would do the same.

The inequalities provided are fairly straightforward given a little thought, and they give you the bound $\delta \leq \frac{1}{10}x_0^3\varepsilon$. But considering our previous argument, we can also require $\delta \leq \frac{1}{2}x_0$, in case $\varepsilon$ is big.