Example 3: Continuity of $f(x)=|x|$ at $x=0$

Basics you need

Solution

1. Write $|x|$ piece by piece

   By the definition of absolute value (Basic 1),

   $$f(x) = |x| = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$$
   Limits are computed from formulas, and $|x|$ is not a single formula: what it does to $x$ depends on the sign of $x$⁠. Splitting it shows exactly which formula applies on each side of 0, the point we are testing.
   Check the left formula on a number: for $x = -2$⁠, $-x = 2 = |-2|$⁠. Basics: piecewise functions

2. Is 0 in the domain? Find $f(0)$

   $|x|$ is defined for every real number, so 0 is in the domain. The value at 0 comes from the line "$x \ge 0$" (because $0 \ge 0$⁠):

   $$f(0) = 0.$$
   At the junction point we must use the piece whose condition includes the point itself. Here both formulas would give 0 at $x = 0$⁠, but in other problems (Example 10) choosing the wrong piece gives a wrong value.
   $|0| = 0$⁠: the distance from 0 to itself is 0.

3. Left-hand limit: use the formula $-x$

   As $x \to 0^-$⁠, every $x$ used is negative, so $f(x) = -x$⁠:

   $$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} (-x) = -0 = 0.$$

   $x$	$f(x)$	side of 0
   $-0.1$	$0.1$	left
   $-0.01$	$0.01$	left
   $-0.001$	$0.001$	left

   The left-hand limit never looks at $x \ge 0$⁠, so the formula $x$ plays no part here. With $-x$⁠: as $x$ climbs to 0 through negative numbers, $-x$ falls to 0 through positive numbers.
   Constant-multiple rule with the constant $-1$⁠: $\lim_{x\to0^-}(-x) = -\lim_{x\to0^-} x$⁠, which is $-0 = 0$⁠. Basics: one-sided limits

4. Right-hand limit: use the formula $x$

   As $x \to 0^+$⁠, every $x$ used is positive, so $f(x) = x$⁠:

   $$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x = 0.$$

   $x$	$f(x)$	side of 0
   $0.1$	$0.1$	right
   $0.01$	$0.01$	right
   $0.001$	$0.001$	right

   Now only $x > 0$ matters, so only the formula $x$ is used. On both sides the distance of $f(x)$ from 0 is $|x|$ itself, which is as small as we like when $x$ is close to 0.

5. Compare the three numbers

   $$\begin{aligned} \lim_{x\to 0^-} f(x) &= 0, \\ \lim_{x\to 0^+} f(x) &= 0, \\ f(0) &= 0. \end{aligned}$$

   The left-hand limit, the right-hand limit and the value coincide, so by Definition 1 $f(x) = |x|$ is continuous at $x = 0$.

   Two different formulas met at 0, yet they arrived at the same height, and that height is $f(0)$⁠. Continuity needs nothing more. The graph has a sharp corner at 0, but a corner is not a break.
   A corner matters for differentiability (Section 5.3), not for continuity: $|x|$ is continuous at 0 but has no derivative there. Lesson: differentiability

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