Example 13: $x$ for $x\ge0$⁠, $x^2$ for $x<0$⁠: continuous everywhere

Basics you need

Solution

1. Find the domain and partition it

   Every real number satisfies exactly one of $x \ge 0$⁠, $x < 0$⁠, so $f$ is defined at every real number: the domain is $\R$⁠. Following the book, split $\R$ into three disjoint sets:

   $$D_1 = \{x \in \R : x < 0\}, \quad D_2 = \{0\},$$ $$D_3 = \{x \in \R : x > 0\}.$$
   The formula changes at 0 only. Points of $D_1$ and $D_3$ lie strictly inside one piece, so a single formula governs all nearby inputs. The point 0 is where two formulas meet, so it gets its own case.
   $D_1 \cup D_2 \cup D_3 = \R$ and the three sets do not overlap, so these three cases examine every real number exactly once. Basics: piecewise functions

2. Case 1: points of $D_1$ ($c < 0$⁠)

   Fix $c < 0$⁠. Then $f(c) = c^2$⁠. Every $x$ with $|x - c| < |c|$ is negative, so near $c$ we have $f(x) = x^2$ and

   $$\lim_{x\to c} f(x) = \lim_{x\to c} x^2 = c^2 = f(c).$$

   So $f$ is continuous at every point of $D_1$.

   Within distance $|c|$ of a negative number $c$ all inputs stay negative, so only the rule $x^2$ is used. The book says "it is easy to see" and points to Example 2 (where $x^2$ is shown continuous at 0); the same limit calculation works at any $c$⁠.
   The product rule of limits: the limit of $x \cdot x$ is the product of the two limits, $$\lim_{x\to c} x^2 = \Big(\lim_{x\to c} x\Big)\Big(\lim_{x\to c} x\Big) = c \cdot c.$$ Basics: what a limit is

3. Case 2: points of $D_3$ ($c > 0$⁠)

   Fix $c > 0$⁠. Then $f(c) = c$⁠. Every $x$ with $|x - c| < c$ is positive, so near $c$ we have $f(x) = x$ and

   $$\lim_{x\to c} f(x) = \lim_{x\to c} x = c = f(c).$$

   So $f$ is continuous at every point of $D_3$.

   Near a positive $c$⁠, $f$ is the identity function, which is continuous at every real number by Example 6.
   Example 6: the identity function is continuous.

4. Case 3: at $x = 0$⁠, the value

   Since $0 \ge 0$⁠, the first line applies:

   $$f(0) = 0.$$
   The condition $x \ge 0$ includes the junction point itself. Using $x^2$ would happen to give the same number here, but only by luck; the correct line is the first.

5. Case 3: the left-hand limit at 0

   For $x < 0$ the rule is $f(x) = x^2$⁠, so

   $$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} x^2 = 0^2 = 0.$$
   Inputs approaching 0 from the left are negative and use the second line. For example $f(-0.1) = 0.01$ and $f(-0.01) = 0.0001$⁠: the outputs shrink to 0, even faster than the inputs.
   Basics: left-hand and right-hand limits

6. Case 3: the right-hand limit at 0

   For $x > 0$ the rule is $f(x) = x$⁠, so

   $$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x = 0.$$
   Inputs approaching 0 from the right are positive and use the first line, where the output equals the input, so the outputs go to 0 as well.

7. Case 3: compare

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

   All three are equal, so $\lim_{x\to 0} f(x) = 0 = f(0)$ and $f$ is continuous at 0.

   Definition 1 asks for exactly this: both one-sided limits exist, agree, and equal the value. In the graph, the parabola and the line meet at the origin with no gap and no jump.
   Lesson: continuity at a point

8. Conclusion

   Cases 1, 2 and 3 cover every real number, and $f$ is continuous in each. So $f$ is continuous at every point of its domain $\R$⁠, and by Definition 2, $f$ is a continuous function.

   The graph can be drawn in one stroke. It does have a sharp turn at the origin: the parabola arrives flat (slope 0) and the line leaves with slope 1. That corner does not affect continuity; it matters for differentiability, which Section 5.3 studies.

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