Example 7: Is $f(x)=|x|$ a continuous function?

Basics you need

Solution

1. Domain, and the three cases

   $|x|$ is defined for every real number, so the domain is $\R$⁠. We rewrite $f$ as

   $$f(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$$

   and examine an arbitrary real number $c$ in three cases: $c < 0$⁠, $c = 0$ and $c > 0$⁠. Every real number falls in exactly one case.

   Definition 2 asks for continuity at every point of the domain. The formula changes at 0, so 0 behaves differently from the other points and must be treated on its own; the negative and positive points each have a single formula nearby.
   Splitting into cases is a valid proof as long as the cases together cover every possibility. Basics: piecewise functions

2. Case $c = 0$⁠: already done

   By Example 3, $\lim_{x\to0^-}|x| = 0$⁠, $\lim_{x\to0^+}|x| = 0$ and $f(0) = 0$⁠. So $f$ is continuous at $x = 0$⁠.

   This is the one point where the left and right formulas differ, so the one-sided limits had to be computed separately with $-x$ and $x$⁠; Example 3 did exactly that.

3. Case $c < 0$⁠: only the formula $-x$ is used near $c$

   Let $c < 0$⁠. Then $f(c) = -c$⁠. Now take any $x$ with $|x - c| < |c|$⁠, that is, any $x$ between $2c$ and 0. Every such $x$ is negative, so $f(x) = -x$ there. Hence

   $$\begin{aligned} \lim_{x\to c^-} f(x) &= \lim_{x\to c^-}(-x) = -c, \\ \lim_{x\to c^+} f(x) &= \lim_{x\to c^+}(-x) = -c. \end{aligned}$$

   Both one-sided limits equal $-c = f(c)$⁠, so $f$ is continuous at $c$⁠, for every $c < 0$⁠.

   This answers the textbook's first "(Why?)": the limit is computed with $-x$ because, close to a negative number $c$⁠, every $x$ is negative as well, and for negative $x$ the rule is $-x$⁠. For $c = -2$⁠, every $x$ in $(-4, 0)$ is negative, and in particular every $x$ in the gold strip of the figure. Then $\lim(-x) = -\lim x = -c$⁠.
   For $c < 0$⁠, $|c| = -c$⁠, so $f(c) = -c$ is a positive number: for $c = -2$⁠, $f(-2) = 2$⁠. Basics: what a limit is

4. Case $c > 0$⁠: only the formula $x$ is used near $c$

   Let $c > 0$⁠. Then $f(c) = c$⁠. Every $x$ with $|x - c| < c$⁠, that is every $x$ between 0 and $2c$⁠, is positive, so $f(x) = x$ there. Hence

   $$\begin{aligned} \lim_{x\to c^-} f(x) &= \lim_{x\to c^-} x = c, \\ \lim_{x\to c^+} f(x) &= \lim_{x\to c^+} x = c. \end{aligned}$$

   Both equal $f(c) = c$⁠, so $f$ is continuous at every $c > 0$⁠.

   This answers the second "(Why?)": near a positive number every $x$ is positive, so $f(x) = x$ there, and $\lim_{x\to c} x = c$ (as in Example 6).
   The window $|x - c| < |c|$ is chosen so that it never reaches 0, the only place where the formula changes. Basics: one-sided limits

5. Put the three cases together

   $f$ is continuous at every $c < 0$ (Step 3), at $c = 0$ (Step 2) and at every $c > 0$ (Step 4). These cases cover all of $\R$⁠, the domain. By Definition 2, $f(x) = |x|$ is a continuous function.

   Definition 2 needs continuity at every point of the domain, and the three cases leave no real number out.
   Lesson: continuity at a point

Every fact used, and where it comes from