Example 11: $x+2$⁠, $0$⁠, $x-2$ around $x=1$⁠: find every discontinuity

Basics you need

Solution

1. Find the domain

   Every real number $x$ satisfies exactly one of $x < 1$⁠, $x = 1$⁠, $x > 1$⁠, and each case has a rule. So $f$ is defined at every real number:

   $$\text{domain of } f = \R.$$

   The question therefore concerns every real number $c$⁠, and we split into three cases: $c < 1$⁠, $c > 1$⁠, $c = 1$⁠.

   Continuity is only ever tested at points of the domain (Definition 1 starts with "let $c$ be a point in the domain"). Here nothing is excluded, so no real number can be skipped. The formula changes only at 1, which is why 1 gets a case of its own.
   The three cases together cover all of $\R$ with no overlap, so after handling them we will have examined every point. Basics: piecewise functions

2. Points $c < 1$

   Fix any $c < 1$⁠. Then $f(c) = c + 2$⁠. The distance from $c$ to 1 is the positive number $1 - c$⁠. Every $x$ with $|x - c| < 1 - c$ satisfies $x < c + (1 - c) = 1$⁠, so on that whole interval around $c$⁠, on both sides of $c$⁠,

   $$f(x) = x + 2.$$

   Hence

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

   So $f$ is continuous at every $c < 1$.

   A limit at $c$ depends only on the values of $f$ at inputs as close to $c$ as we like. Staying within distance $1 - c$ of $c$ we never reach 1, so the lines "$0$" and "$x - 2$" are never used near $c$⁠: near $c$⁠, $f$ simply is the line $x + 2$⁠. For example, at $c = 0.9$ every $x$ in $(0.8, 1)$ is below 1.
   $\lim_{x\to c}(x + 2) = c + 2$ by the algebra of limits (Basic 3). Basics: what a limit is

3. Points $c > 1$

   Fix any $c > 1$⁠. Then $f(c) = c - 2$⁠. Every $x$ with $|x - c| < c - 1$ satisfies $x > c - (c - 1) = 1$⁠, 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 $c > 1$.

   The mirror image of Step 2: a small enough interval around $c$ lies entirely to the right of 1, where only the rule $x - 2$ is used. This is what the book means by "as in the previous example we find that $f$ is continuous at all real numbers $x \ne 1$".
   Example 10 uses exactly the same two lines away from 1.

4. At $c = 1$⁠: the value

   The input 1 satisfies only the condition $x = 1$⁠, so the middle line applies:

   $$f(1) = 0.$$
   It is tempting to put $x = 1$ into $x + 2$ or $x - 2$⁠, but those rules are only for $x < 1$ and $x > 1$⁠. The value at 1 is fixed separately by the definition, and it is 0.
   Basics: piecewise functions

5. At $c = 1$⁠: the left-hand limit

   For inputs $x < 1$ the rule is $f(x) = x + 2$⁠, so

   $$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}(x + 2) = 1 + 2 = 3.$$
   "$x \to 1^-$" means $x$ approaches 1 through numbers less than 1, such as 0.9, 0.99, 0.999. All of them use the first line. The value $f(1) = 0$ plays no part, because a limit never uses the value at the point itself.
   Basics: left-hand and right-hand limits

6. At $c = 1$⁠: the right-hand limit

   For inputs $x > 1$ the rule is $f(x) = x - 2$⁠, so

   $$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+}(x - 2) = 1 - 2 = -1.$$

   $x$	line used	$f(x)$
   $0.9$	$x + 2$	$2.9$
   $0.99$	$x + 2$	$2.99$
   $0.999$	$x + 2$	$2.999$
   $1$	$0$	$0$
   $1.001$	$x - 2$	$-0.999$
   $1.01$	$x - 2$	$-0.99$
   $1.1$	$x - 2$	$-0.9$

   Inputs just above 1 use the third line, and $x - 2$ heads for $1 - 2 = -1$⁠. The table shows the three behaviours side by side: values near 3 on the left, the single value 0 at 1, values near $-1$ on the right.

7. Compare the three numbers

   $$\begin{aligned} \lim_{x\to 1^-} f(x) &= 3, \\ \lim_{x\to 1^+} f(x) &= -1, \\ f(1) &= 0. \end{aligned}$$

   The left-hand and right-hand limits are different ($3 \ne -1$⁠), so $\lim_{x\to 1} f(x)$ does not exist, and Definition 1 cannot hold. Hence $f$ is not continuous at $x = 1$.

   Continuity at 1 needs all three numbers to be equal. The two one-sided limits already disagree, so the test fails before $f(1)$ is even compared. (The value 0 also matches neither limit, but that is a second, separate defect.) The graph jumps by $3 - (-1) = 4$ at $x = 1$⁠.
   Because the one-sided limits differ, no choice of the single value $f(1)$ could make $f$ continuous at 1: 0, 3 or $-1$ would all fail. Lesson: continuity at a point

8. Conclusion

   By Steps 2 and 3, $f$ is continuous at every real number other than 1; by Step 7 it is not continuous at 1. So $x = 1$ is the only point of discontinuity of $f$.

   Steps 1 to 3 and 7 cover every real number, so no other point of discontinuity can exist. In the graph, the pen must be lifted at $x = 1$ and only there.

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