Example 10: $x+2$ for $x\le1$⁠, $x-2$ for $x>1$⁠: a jump at 1

Basics you need

Solution

1. The domain and the cases

   Every real number satisfies exactly one of $x \le 1$⁠, $x > 1$⁠, so $f$ is defined at all points of the real line. We test an arbitrary real number $c$ in three cases: $c < 1$⁠, $c > 1$ and $c = 1$⁠.

   The formula changes only at 1, so 1 is the only point where the two sides of a point can use different formulas. Every other point lies strictly inside one piece.
   Basics: piecewise functions

2. Case 1: $c < 1$

   Here $f(c) = c + 2$⁠. Every $x$ with $|x - c| < 1 - c$ satisfies $x < 1$⁠, so near $c$⁠, on both sides, $f(x) = x + 2$⁠. Hence

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

   So $f$ is continuous at every real number less than 1.

   Within distance $1 - c$ of $c$ we cannot reach 1, so only the formula $x + 2$ is ever used. A limit only looks near $c$⁠, so it is the limit of the line $x + 2$⁠.
   For $c = 0.9$⁠: every $x$ in $(0.8, 1)$ is less than 1. Basics: what a limit is

3. Case 2: $c > 1$

   Here $f(c) = c - 2$⁠. Every $x$ with $|x - c| < c - 1$ satisfies $x > 1$⁠, so near $c$ we have $f(x) = x - 2$⁠, and

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

   So $f$ is continuous at every point $x > 1$⁠.

   Same reasoning on the other side: a small enough interval around $c$ stays to the right of 1, where only $x - 2$ is used.

4. Case 3: $c = 1$⁠, the value

   Since $1 \le 1$⁠, the first line applies:

   $$f(1) = 1 + 2 = 3.$$
   The condition "$x \le 1$" includes the junction point; "$x > 1$" does not. Using $x - 2$ here would give $-1$⁠, which is not the value of $f$ at 1.
   Basics: piecewise functions

5. Case 3: left-hand limit at 1

   For $x < 1$⁠, $f(x) = x + 2$⁠:

   $$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}(x + 2) = 1 + 2 = 3.$$
   Numbers just below 1 obey the first line, so the left-hand limit is computed with $x + 2$ only.
   Basics: one-sided limits

6. Case 3: right-hand limit at 1

   For $x > 1$⁠, $f(x) = x - 2$⁠:

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

   $x$	$f(x)$	which line
   $0.9$	$2.9$	$x \le 1$
   $0.99$	$2.99$	$x \le 1$
   $1$	$3$	$x \le 1$
   $1.01$	$-0.99$	$x > 1$
   $1.1$	$-0.9$	$x > 1$
   $1.5$	$-0.5$	$x > 1$

   Numbers just above 1 obey the second line, and $x - 2$ heads for $1 - 2 = -1$⁠. The table shows the values on the right staying near $-1$⁠, far from the values on the left near 3.

7. Case 3: compare

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

   The left-hand and right-hand limits do not coincide ($3 \ne -1$⁠), so $\lim_{x\to1} f(x)$ does not exist, and Definition 1 cannot hold. $f$ is not continuous at $x = 1$.

   The value $f(1) = 3$ does match the left-hand limit, but that is not enough: continuity needs all three numbers to agree. The graph jumps by $3 - (-1) = 4$ at $x = 1$⁠, so it cannot be drawn there without lifting the pen.
   Lesson: continuity at a point

8. Conclusion

   By Cases 1 and 2, $f$ is continuous at every $c \ne 1$⁠; by Case 3 it is not continuous at 1. So $x = 1$ is the only point of discontinuity of $f$.

   The three cases cover every real number, so no other point can be a point of discontinuity.

Every fact used, and where it comes from