Exercise 5.1 · Q14: Continuity of the step function 3, 4, 5 on $[0, 10]$

Basics you need

Solution

1. Find the domain and the special points

   The three conditions cover $[0, 1]$⁠, $(1, 3)$ and $[3, 10]$⁠. These do not overlap ($1$ belongs only to the first, $3$ only to the third) and together they make up $[0, 10]$⁠. So the domain of $f$ is $[0, 10]$⁠, and $f$ is not defined for $x < 0$ or $x > 10$⁠.

   The special points are the end points $0$⁠, $10$ and the junctions $1$⁠, $3$⁠.

   Continuity is only discussed at points of the domain. A number like $-2$ or $11$ is neither a point of continuity nor a point of discontinuity of $f$⁠; $f$ simply does not exist there. And at the end points of a closed interval the test is different, so we must spot them first.
   Read the inequalities carefully: $\le$ includes the end number, $<$ excludes it. Basics: piecewise functions

2. Points strictly inside a piece

   If $0 < c < 1$⁠: all $x$ within distance $\min(c, 1 - c)$ of $c$ lie in $(0, 1)$⁠, where $f(x) = 3$⁠. So $\lim_{x\to c} f(x) = 3 = f(c)$⁠.

   If $1 < c < 3$⁠: all $x$ within distance $\min(c - 1, 3 - c)$ of $c$ lie in $(1, 3)$⁠, where $f(x) = 4$⁠. So $\lim_{x\to c} f(x) = 4 = f(c)$⁠.

   If $3 < c < 10$⁠: all $x$ within distance $\min(c - 3, 10 - c)$ of $c$ lie in $(3, 10)$⁠, where $f(x) = 5$⁠. So $\lim_{x\to c} f(x) = 5 = f(c)$⁠.

   So $f$ is continuous at every point of $(0, 1)$⁠, $(1, 3)$ and $(3, 10)$⁠.

   Around such a point there is a small interval on which $f$ is one constant. The limit only looks at that small interval, and the limit of a constant is the constant. The distance $\min(\dots)$ is just the distance to the nearer special point, so the small interval does not reach a junction or an end.
   Constant functions are continuous (Example 5). Lesson: continuity at a point

3. The end points $x = 0$ and $x = 10$

   At $x = 0$⁠. $f(0) = 3$ (since $0 \le 0 \le 1$⁠). There are no points of the domain to the left of $0$⁠, so only the right-hand limit is used. For $0 < x < 1$⁠, $f(x) = 3$⁠:

   $$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} 3 = 3 = f(0).$$

   So $f$ is continuous at $x = 0$⁠.

   At $x = 10$⁠. $f(10) = 5$ (since $3 \le 10 \le 10$⁠). There are no points of the domain to the right of $10$⁠, so only the left-hand limit is used. For $3 < x < 10$⁠, $f(x) = 5$⁠:

   $$\lim_{x\to 10^-} f(x) = \lim_{x\to 10^-} 5 = 5 = f(10).$$

   So $f$ is continuous at $x = 10$⁠.

   The chapter's elaboration of Definition 2 says exactly this: for a function on $[a, b]$⁠, continuity at $a$ means $\lim_{x\to a^+} f(x) = f(a)$ and at $b$ means $\lim_{x\to b^-} f(x) = f(b)$⁠. The limits $\lim_{x\to 0^-}$ and $\lim_{x\to 10^+}$ "do not make sense", in the book's words, because $f$ has no values there. It would be wrong to call $0$ or $10$ discontinuities just because the graph stops.
   At an end point of a closed interval, use only the one-sided limit from inside the interval. Basics: one-sided limits

4. The junction $x = 1$

   Value. $1$ satisfies $0 \le x \le 1$⁠, so $f(1) = 3$⁠.

   Left-hand limit. For $0 \le x < 1$⁠, $f(x) = 3$⁠:

   $$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} 3 = 3.$$

   Right-hand limit. For $1 < x < 3$⁠, $f(x) = 4$⁠:

   $$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} 4 = 4.$$

   Compare. $3 \ne 4$⁠, so the limit at $1$ does not exist and $f$ is discontinuous at $x = 1$.

   The value $f(1) = 3$ matches the left side, but the right side is at height $4$⁠. A jump of $1$ means the one-sided limits disagree, which already rules out continuity.
   Unequal one-sided limits: no limit, so no continuity.

5. The junction $x = 3$

   Value. $3$ satisfies $3 \le x \le 10$⁠, so $f(3) = 5$⁠.

   Left-hand limit. For $1 < x < 3$⁠, $f(x) = 4$⁠:

   $$\lim_{x\to 3^-} f(x) = \lim_{x\to 3^-} 4 = 4.$$

   Right-hand limit. For $3 < x \le 10$⁠, $f(x) = 5$⁠:

   $$\lim_{x\to 3^+} f(x) = \lim_{x\to 3^+} 5 = 5.$$

   Compare. $4 \ne 5$⁠, so $f$ is discontinuous at $x = 3$.

   This time the value $f(3) = 5$ matches the right side, but the left side is at height $4$⁠. Whichever side the junction value belongs to, a jump means discontinuity.
   Definition 1 needs both one-sided limits and the value to agree. Lesson: continuity at a point

6. Conclude

   Steps 2 to 5 cover every point of the domain $[0, 10]$⁠. $f$ is continuous at every point of $[0, 10]$ except $x = 1$ and $x = 3$⁠, where it is discontinuous. So the points of discontinuity are $x = 1$ and $x = 3$, and $f$ is not a continuous function.

   "Discuss the continuity" asks for the full picture: where $f$ is continuous, where it is not, and why. The list "inside the pieces, the two end points, the two junctions" accounts for every $x$ with $0 \le x \le 10$⁠.

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