Exercise 5.2 · Q10: $[x]$ on $(0,3)$ is not differentiable at $x=1$ and $x=2$

Basics you need

Solution

1. Write $f$ step by step on $(0, 3)$

   $$f(x) = [x] = \begin{cases} 0, & 0 < x < 1 \\ 1, & 1 \le x < 2 \\ 2, & 2 \le x < 3 \end{cases}$$

   Both $1$ and $2$ are in the domain, with $f(1) = 1$ and $f(2) = 2$⁠.

   Differentiability is only asked at points of the domain, so we first confirm that $1$ and $2$ belong to $(0, 3)$⁠. Each integer belongs to the step that starts at it, because $[1] = 1$⁠, not $0$⁠.
   $[x] = n$ exactly when $n \le x < n + 1$ ($n$ an integer). Basics: piecewise functions

2. At $x = 1$⁠: right-hand derivative

   Take $0 < h < 1$⁠. Then $1 < 1 + h < 2$⁠, so $f(1 + h) = 1$ and

   $$\frac{f(1+h) - f(1)}{h} = \frac{1 - 1}{h} = 0.$$

   Hence $\displaystyle\lim_{h\to0^+}\frac{f(1+h) - f(1)}{h} = 0$⁠.

   To the right of $1$ we stay on the flat step at height $1$⁠, and $f(1)$ is also $1$⁠, so every chord is horizontal. Restricting to $h < 1$ keeps $1 + h$ on that step (and inside the domain); only small $h$ matter for the limit anyway.
   A quotient that equals $0$ for every small $h > 0$ has right-hand limit $0$⁠. Basics: one-sided limits

3. At $x = 1$⁠: left-hand difference quotient

   Take $-1 < h < 0$⁠. Then $0 < 1 + h < 1$⁠, so $f(1 + h) = 0$ and

   $$\frac{f(1+h) - f(1)}{h} = \frac{0 - 1}{h} = -\frac{1}{h}.$$

   Since $h < 0$⁠, $-\frac1h = \frac{1}{|h|}$⁠, which is positive and grows without bound as $h \to 0^-$⁠:

   $$\begin{aligned} h &= -0.1,\ -0.01,\ -0.001 \\ -\tfrac1h &= 10,\ 100,\ 1000. \end{aligned}$$

   So $\displaystyle\lim_{h\to0^-}\frac{f(1+h) - f(1)}{h}$ is not a finite number: the left-hand derivative at $1$ does not exist.

   Just left of $1$ the graph is on the lower step (height $0$⁠), while $f(1) = 1$ is on the upper step. The chord from $(1+h, 0)$ to $(1, 1)$ rises by $1$ over a horizontal distance $|h|$⁠, so its slope is $\frac{1}{|h|}$⁠, which becomes as large as we like.
   $\frac{1}{|h|}$ exceeds any number $M$ once $|h| < \frac1M$⁠, so it has no finite limit. Basics: what a limit is

4. Conclusion at $x = 1$

   The definition needs both one-sided limits to be finite and equal. The left-hand one is not finite (and the right-hand one is $0$⁠), so $f$ is not differentiable at $x = 1$.

   One failed condition is enough. Even if we tried to call the left-hand value "$+\infty$", it would neither be finite nor equal to the right-hand value $0$⁠.
   Definition of differentiability at a point, Section 5.3. Lesson: differentiability

5. At $x = 2$⁠: the same two calculations

   Right. For $0 < h < 1$⁠: $2 < 2 + h < 3$⁠, so $f(2 + h) = 2$ and

   $$\frac{f(2+h) - f(2)}{h} = \frac{2 - 2}{h} = 0 \to 0.$$

   Left. For $-1 < h < 0$⁠: $1 < 2 + h < 2$⁠, so $f(2 + h) = 1$ and

   $$\frac{f(2+h) - f(2)}{h} = \frac{1 - 2}{h} = -\frac1h = \frac{1}{|h|},$$

   which grows without bound as $h \to 0^-$⁠. The left-hand derivative at $2$ does not exist, so $f$ is not differentiable at $x = 2$.

   The situation at $2$ is a copy of the one at $1$⁠, one step higher: flat to the right, a jump of $1$ to the left. Note $2 + h < 3$ keeps us inside the domain $(0, 3)$⁠.

6. A second proof, through Theorem 3

   At $x = 1$⁠: for $x$ just left of $1$⁠, $f(x) = 0$⁠; just right, $f(x) = 1$⁠. So

   $$\lim_{x\to1^-} f(x) = 0 \ne 1 = \lim_{x\to1^+} f(x),$$

   and $f$ is not continuous at $1$⁠. Likewise $\lim_{x\to2^-} f(x) = 1$ while $\lim_{x\to2^+} f(x) = 2$⁠. If $f$ were differentiable at $1$ or $2$⁠, Theorem 3 would make it continuous there, which is false. So $f$ is not differentiable at $1$ or $2$⁠.

   This is the contrapositive of Theorem 3: "differentiable $\Rightarrow$ continuous" is the same statement as "not continuous $\Rightarrow$ not differentiable". The book shows in Example 15 that $[x]$ is discontinuous at every integer; restricted to $(0, 3)$⁠, the integers in the domain are exactly $1$ and $2$⁠.
   Theorem 3, Section 5.3; Example 15, Section 5.2. Lesson: differentiability

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