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

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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.

Oscar Flores
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2 Answers2

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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.

Henry
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  • What would be the issue if instead of $x_0$, you chose an arbitrary positive real number to bound $|x-x_0|$? – Oscar Flores Nov 13 '14 at 23:41
  • If your number was $k$ then $x_0$ might be so small that $0 \lt x_0 \lt k$ and so $|x-x_0| \lt k$ might still allow $x$ to be negative, zero or arbitrarily small. The only way to avoid the problem would be scale $k$ to $x_0$. – Henry Nov 13 '14 at 23:48
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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.

neptun
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