1. 미분 (i): 극한·연속·도함수Differentiation (i): Limits, Continuity, and the Derivative

We simply transfer the definition of the right limit in Definition 1.6(ii) to the left side. Let $a, b$ be real numbers such that $a \lt b$ and $f$ be a real-valued function defined on the open interval $(a, b)$.

For $d \in \mathbb{R}$ and $c \in (a, b]$, we say that $f$ has a left limit $d$ at $c$ iff for all $\varepsilon > 0$, there exists a $\delta > 0$ such that

for all $x \in (c - \delta, c)$. For the left limit, we use the notation

or

Symmetrically to Definition 1.6(ii), which defines the right limit for $c \in [a, b)$, the left limit is defined for $c \in (a, b]$. Indeed, if $c \in (a, b]$ and $\delta \le c - a$, then $(c - \delta, c) \subset (a, b)$, so the condition above is meaningful; as in Definition 1.6(ii), there is no loss of generality in taking $\delta$ that small.

The domain $(a, b)$ of $f$ is a metric subspace of $\mathbb{R}$ (Remark 1.2) and $c \in (a, b]$ is a limit point of $(a, b)$, so the limit of $f$ at $c$ is also defined by Definition 1.3. If $\delta \le c - a$, then

and in particular, when $c = b$ the second set on the right is empty, so the set above equals $(b - \delta, b)$. Hence, symmetrically to Remark 1.7,

Let $c \in (a, b)$ and let $\varepsilon > 0$ be given. Any $\delta > 0$ will do, but so that the condition in Definition 1.6(i) is meaningful inside the domain of $f$, put

Then $(c - \delta, c) \cup (c, c + \delta) \subset (a, b)$, so for any element $x$ of this set we have $f(x) = d$ and hence

Since $\varepsilon > 0$ was arbitrary, Definition 1.6(i) gives

Note that $\delta$ does not depend on $\varepsilon$ at all. The same computation gives the one-sided limits as well, provided $\delta$ is taken one-sided. Indeed, for $c \in [a, b)$ put $\delta := b - c > 0$; then $(c, c + \delta) \subset (a, b)$, so $|f(x) - d| = 0 \lt \varepsilon$ at every $x$ of that set and Definition 1.6(ii) gives $\lim_{x \to c^+} f(x) = d$. Likewise, for $c \in (a, b]$ put $\delta := c - a > 0$; then $(c - \delta, c) \subset (a, b)$ and the definition of Exercise 1.8 gives $\lim_{x \to c^-} f(x) = d$. (At an endpoint $c = a$ or $c = b$ one has $\min\{c - a, \ b - c\} = 0$, so the $\delta$ used above is unavailable and only the relevant one-sided width may be taken.)

In the statement, the equality $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ is read as: both one-sided limits exist and their values agree. For this to be meaningful the value of a limit must be unique, which is checked as follows. If $d_1 \neq d_2$ were both limits of $f$ at $c$ (or both right limits, or both left limits), then for $\varepsilon := |d_1 - d_2| / 2 > 0$ we obtain $\delta_1, \delta_2 > 0$, and taking the corresponding (one-sided) neighbourhood with $\delta := \min\{\delta_1, \delta_2\}$, which is nonempty, any point $x$ in it satisfies

a contradiction. Hence the value of a limit is unique.

($\Longrightarrow$) Assume $\lim_{x \to c} f(x) = d$ exists. Let $\varepsilon > 0$ be given. By Definition 1.6(i) there exists a $\delta > 0$ such that

In particular $|f(x) - d| \lt \varepsilon$ for all $x \in (c, c + \delta)$, so $\lim_{x \to c^+} f(x) = d$ by Definition 1.6(ii); and in particular $|f(x) - d| \lt \varepsilon$ for all $x \in (c - \delta, c)$, so $\lim_{x \to c^-} f(x) = d$ by the definition in Exercise 1.8. Thus both one-sided limits exist and both equal $d$.

($\Longleftarrow$) Assume both one-sided limits exist and let $d$ be their common value. Let $\varepsilon > 0$ be given. By the definition in Exercise 1.8 there exists a $\delta_1 > 0$ such that

and by Definition 1.6(ii) there exists a $\delta_2 > 0$ such that

Now put

Let $x \in (c - \delta, c) \cup (c, c + \delta)$. There are two cases. If $x \in (c - \delta, c)$, then $(c - \delta, c) \subset (c - \delta_1, c)$ and hence $|f(x) - d| \lt \varepsilon$; if $x \in (c, c + \delta)$, then $(c, c + \delta) \subset (c, c + \delta_2)$ and again $|f(x) - d| \lt \varepsilon$. In either case $|f(x) - d| \lt \varepsilon$, so since $\varepsilon > 0$ was arbitrary, Definition 1.6(i) gives

in particular the limit of $f$ at $c$ exists.

Combining the two directions, $\lim_{x \to c} f(x)$ exists if and only if $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ both exist and are equal, and in that case all three values coincide.

Definition 1.6(i) is the special case of Definition 1.3 obtained by taking $X = Y = \mathbb{R}$, $E = (a, b)$, $p = c$, $q = d$, and the metric of Example 1.5 with $k = 1$, that is, $d(x, y) = |x - y|$ from (1.1). Indeed the two conditions match as follows.

(The conclusion.) The conclusion in Definition 1.3 is $d_Y(f(x), q) \lt \varepsilon$; since $Y = \mathbb{R}$ and $d_Y(u, v) = |u - v|$, this is $|f(x) - d| \lt \varepsilon$, the same inequality as the conclusion of Definition 1.6(i).

(The range of $x$.) In Definition 1.3, $x$ ranges over all elements of $E$ satisfying $0 \lt d_X(x, p) \lt \delta$. Since $d_X(x, c) = |x - c|$,

As $c \in (a, b)$, choosing $\delta \le \min\{c - a, \ b - c\}$ makes the right-hand side exactly $(c - \delta, c) \cup (c, c + \delta)$, the set over which $x$ ranges in Definition 1.6(i). Shrinking $\delta$ never weakens the condition (the condition for a smaller $\delta$ follows from the one for a larger $\delta$), so whichever definition is used, the set of admissible limits is the same.

In particular both definitions exclude the point $c$ itself: in Definition 1.3 this is done by the left inequality $0 \lt d_X(x, p)$, i.e. $x \neq c$, and in Definition 1.6(i) by writing the neighbourhood as $(c - \delta, c) \cup (c, c + \delta)$ rather than $(c - \delta, c + \delta)$. Hence the value $f(c)$ (indeed, whether it is defined at all) is irrelevant to the limit.

The differences lie in form and in generality.

(a) Definition 1.3 explicitly demands $q \in Y$. In Definition 1.6(i) one starts with $d \in \mathbb{R} = Y$, so this demand is automatically met.

(b) Definition 1.3 requires $p$ to be a limit point of $E$. In Definition 1.6(i) the set $E = (a, b)$ is an open interval and $c \in (a, b)$, so $c$ is always a limit point of $E$ and this requirement is automatic as well. (Without it, at an isolated point every $q$ would be a limit and uniqueness of the limit would fail.)

(c) Definition 1.3 works for arbitrary metric spaces $X, Y$ and an arbitrary subset $E \subset X$. For instance it applies verbatim to vector-valued functions with $Y = \mathbb{R}^k$, or to a domain containing an endpoint such as $E = [a, b)$ (Remark 1.2, Remark 1.7). Definition 1.6(i), by contrast, covers only one situation: a real-valued function on an open interval of $\mathbb{R}$.

(d) The reason Definition 1.6 is nevertheless needed is part (ii), the right limit. A one-sided limit, in which $x$ is restricted to $(c, c + \delta)$, is not directly expressible on $(a, b)$ in the form of Definition 1.3; to phrase it through Definition 1.3 one must first restrict $f$ to $(c, b)$ and then take the limit at $c$, and Remark 1.7 is exactly an instance of that fact.

In summary, on $(a, b)$ with the Euclidean metric, Definition 1.6(i) is a verbatim translation of Definition 1.3, and the two definitions determine the same $\lim_{x \to c} f(x)$.

Let $c \in (a, b)$ and assume that

exist. To shorten the notation, for $0 \lt \delta \le \min\{c - a, \ b - c\}$ write

Definition 1.6(i) has the form "for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - L| \lt \varepsilon$ on $N_\delta$", and we repeatedly use that $N_{\delta_1} \cap N_{\delta_2} = N_{\min\{\delta_1, \delta_2\}}$.

(A lemma: local boundedness.) If $\lim_{x \to c} f(x) = L$, then there exists a $\delta_0 > 0$ such that

Indeed, taking $\varepsilon = 1$ in Definition 1.6(i) gives a $\delta_0 > 0$ with $|f(x) - L| \lt 1$ on $N_{\delta_0}$, and by the triangle inequality $|f(x)| \le |f(x) - L| + |L| \lt |L| + 1$.

(i) Let $\varepsilon > 0$ be given. Applying Definition 1.6(i) to $f$ and to $g$ with $\varepsilon / 2 > 0$ gives $\delta_1, \delta_2 > 0$ such that

Put $\delta := \min\{\delta_1, \delta_2\}$. Then for all $x \in N_\delta$, by the triangle inequality,

Since $\varepsilon > 0$ was arbitrary, $\lim_{x \to c} (f + g)(x) = L + M$.

(ii) Adding and subtracting a term, for all $x \in (a, b)$,

Let $\varepsilon > 0$ be given. Take $\delta_0 > 0$ from the lemma, so that $|f(x)| \lt |L| + 1$ on $N_{\delta_0}$, and use Definition 1.6(i) to choose $\delta_1, \delta_2 > 0$ with

(both denominators are positive). Put $\delta := \min\{\delta_0, \delta_1, \delta_2\}$. Then for all $x \in N_\delta$,

Hence $\lim_{x \to c} (fg)(x) = LM$.

(iii) Assume $M \neq 0$. We first show that

Taking $\varepsilon = |M| / 2 > 0$ in Definition 1.6(i) gives a $\delta_3 > 0$ such that

Then for all $x \in N_{\delta_3}$, by the triangle inequality,

so in particular $g(x) \neq 0$ on $N_{\delta_3}$ and both $1/g$ and $f/g$ are defined there. Moreover, on that range,

Now let $\varepsilon > 0$ be given. By Definition 1.6(i) there is a $\delta_4 > 0$ with $|g(x) - M| \lt \dfrac{|M|^2}{2}\varepsilon$ on $N_{\delta_4}$. Putting $\delta := \min\{\delta_3, \delta_4\}$, for all $x \in N_\delta$ we get

(the inequality is strict because the last one used is strict). Hence $\lim_{x \to c} 1/g(x) = 1/M$. Finally, on $N_{\delta_3}$ we have $\dfrac{f}{g}(x) = f(x) \cdot \dfrac{1}{g(x)}$, so part (ii), already proved, gives

The proof in Exercise 1.13 uses only the following two properties of the set $N_\delta = (c - \delta, c) \cup (c, c + \delta)$ that governs the limit.

(P1) The definition has the form "for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - L| \lt \varepsilon$ at every point of that set".

(P2) The intersection of two such sets is again a set of the same shape: with $\delta := \min\{\delta_1, \delta_2\}$ we have $N_{\delta_1} \cap N_{\delta_2} = N_\delta$.

The set $R_\delta := (c, c + \delta)$ governing the right limit and the set $L_\delta := (c - \delta, c)$ governing the left limit satisfy (P1) and (P2) as well; indeed

Therefore the argument of Exercise 1.13 goes through word for word once $N_\delta$ is replaced by $R_\delta$ or $L_\delta$ and $\lim_{x \to c}$ by $\lim_{x \to c^+}$ or $\lim_{x \to c^-}$. Below we write the right-limit case out in full and note that the left-limit case is symmetric.

(Right limits.) Let $c \in [a, b)$ and assume that

exist. For $0 \lt \delta \le b - c$ we have $R_\delta = (c, c + \delta) \subset (a, b)$.

(Local boundedness.) Taking $\varepsilon = 1$ in Definition 1.6(ii) gives a $\delta_0 > 0$ with $|f(x) - L| \lt 1$ on $R_{\delta_0}$, hence

(i) Let $\varepsilon > 0$ be given. Definition 1.6(ii) gives $\delta_1, \delta_2 > 0$ with $|f(x) - L| \lt \varepsilon / 2$ on $R_{\delta_1}$ and $|g(x) - M| \lt \varepsilon / 2$ on $R_{\delta_2}$. Putting $\delta := \min\{\delta_1, \delta_2\}$, for all $x \in R_\delta$,

that is, $\lim_{x \to c^+} (f + g)(x) = L + M$.

(ii) Let $\varepsilon > 0$ be given. Together with $\delta_0$ above, choose $\delta_1, \delta_2 > 0$ with $|g(x) - M| \lt \dfrac{\varepsilon}{2(|L| + 1)}$ on $R_{\delta_1}$ and $|f(x) - L| \lt \dfrac{\varepsilon}{2(|M| + 1)}$ on $R_{\delta_2}$, and put $\delta := \min\{\delta_0, \delta_1, \delta_2\}$. Then for all $x \in R_\delta$,

that is, $\lim_{x \to c^+} (fg)(x) = LM$.

(iii) Assume $M \neq 0$. Taking $\varepsilon = |M| / 2$ in Definition 1.6(ii) gives a $\delta_3 > 0$ with $|g(x) - M| \lt |M| / 2$ on $R_{\delta_3}$, so that there

Given $\varepsilon > 0$, choose $\delta_4 > 0$ with $|g(x) - M| \lt \dfrac{|M|^2}{2}\varepsilon$ on $R_{\delta_4}$ and put $\delta := \min\{\delta_3, \delta_4\}$; then $\left| 1/g(x) - 1/M \right| \lt \varepsilon$ on $R_\delta$. Thus $\lim_{x \to c^+} 1/g(x) = 1/M$, and applying part (ii), already proved, to $f$ and $1/g$,

(Left limits.) In the argument above replace $R_\delta = (c, c + \delta)$ by $L_\delta = (c - \delta, c)$, the condition $c \in [a, b)$ by $c \in (a, b]$, the restriction $\delta \le b - c$ by $\delta \le c - a$, and $\lim_{x \to c^+}$ by $\lim_{x \to c^-}$ (whose definition is that of Exercise 1.8). Not a single inequality or choice of $\delta$ changes, precisely because the argument depends only on (P1) and (P2). Hence formulas (i), (ii), and (iii) of Theorem 1.12 hold for left limits and right limits.

Let $f$ be a real-valued function defined on the closed interval $[a, b]$.

(i) Cutting the neighbourhood $(c - \delta, c + \delta) \cap [a, b]$ of Definition 1.15 to one side gives the following definitions.

For $c \in [a, b)$, we say that $f$ is right continuous at $c$ iff for all $\varepsilon > 0$ there exists a $\delta > 0$ such that

For $c \in (a, b]$, we say that $f$ is left continuous at $c$ iff for all $\varepsilon > 0$ there exists a $\delta > 0$ such that

At $x = c$ the inequality $|f(c) - f(c)| = 0 \lt \varepsilon$ always holds, so deleting the point $c$ from the two conditions changes nothing. Hence, for $c \in [a, b)$, $f$ is right continuous at $c$ if and only if $\lim_{x \to c^+} f(x) = f(c)$, and for $c \in (a, b]$, $f$ is left continuous at $c$ if and only if $\lim_{x \to c^-} f(x) = f(c)$ (Definition 1.6(ii) and Exercise 1.8). In particular, for $c \in (a, b)$, Exercise 1.10 shows that $f$ is continuous at $c$ if and only if $f$ is both left and right continuous at $c$.

(ii) At $c = a$ the range of $x$ required by Definition 1.15 is

(the inequality $x > a - \delta$ is automatic because $x \ge a$). But this is literally the same set as the range of $x$ required by right continuity in (i) with $c = a$. The conclusions ($|f(x) - f(a)| \lt \varepsilon$) and the order of the quantifiers agree as well, so $f$ is continuous at $a$ if and only if $f$ is right continuous at $a$. Symmetrically, at $c = b$,

so $f$ is continuous at $b$ if and only if $f$ is left continuous at $b$. In other words, at an endpoint the notion of "continuity" in Definition 1.15 silently weakens to one-sided continuity. This is the starting point of the trouble discussed in Exercise 1.23.

(iii) Apply Definition 1.4 to the following data: $X = [a, b]$, $Y = \mathbb{R}$, $E = [a, b]$, $p = c$, and the metric $d(u, v) = |u - v|$ of (1.1). By Remark 1.2 the set $[a, b] \subset \mathbb{R}$ is a metric space with this metric, so the application is legitimate. Definition 1.4 then says

Definition 1.15, on the other hand, says

But for every $\delta > 0$,

Indeed, for $x \in [a, b]$ the inequality $|x - c| \lt \delta$ is equivalent to $c - \delta \lt x \lt c + \delta$, so the two sets have the same elements. The two definitions are therefore the same statement: same $\varepsilon$, same $\delta$, same conclusion, same range of $x$. Hence, whether $c$ is an interior point or an endpoint, $f$ is continuous at $c$ in the sense of Definition 1.4 if and only if it is continuous at $c$ in the sense of Definition 1.15. That is, Definition 1.15 is nothing but Definition 1.4 written out in the case $X = E = [a, b]$, $Y = \mathbb{R}$.

(i) Let $f$ be a real-valued function defined on the closed interval $[a, b]$. As in Definition 1.17, fix $x \in [a, b]$ and put

Replacing the limit $\lim_{t \to x} \phi(t)$ of Definition 1.17 by a one-sided limit gives the following definitions.

For $x \in [a, b)$, if the limit

exists, its value is called the right derivative of $f$ at $x$, and we say that $f$ is right differentiable at $x$.

For $x \in (a, b]$, if the limit

exists, its value is called the left derivative of $f$ at $x$, and we say that $f$ is left differentiable at $x$.

Here the right limit is in the sense of Definition 1.6(ii) and the left limit in the sense of Exercise 1.8. Since the domain of $\phi$ is $\{t \in (a, b) : t \neq x\}$, one cannot speak of a right limit at $x = b$ nor of a left limit at $x = a$; this is why the definitions are restricted to $x \in [a, b)$ and $x \in (a, b]$ respectively.

(ii) Definition 1.17 defines the derivative at $x = a$ in the first place by

which is literally the same expression as the $f'_+(a)$ defined in (i). Hence the right derivative at $a$ and the derivative at $a$ are not two objects that happen to have the same value; they are the same object by definition:

One exists if and only if the other does, and then the values agree. Symmetrically, $f'_-(b) = f'(b)$ at $x = b$.

At an interior point $x \in (a, b)$ the situation is different, because there Definition 1.17 requires the two-sided limit $\lim_{t \to x} \phi(t)$. Applying Exercise 1.10 to $\phi$, the derivative $f'(x)$ exists if and only if $f'_-(x)$ and $f'_+(x)$ both exist and are equal, and then all three values coincide. For instance $f(x) = |x|$ of Example 1.34(i) has $f'_+(0) = 1$ and $f'_-(0) = -1$ at $x = 0$, so both one-sided derivatives exist but differ, and $f'(0)$ does not exist.

In summary, at an endpoint the "derivative" of Definition 1.17 is exactly a one-sided derivative (which is why, as in Remark 1.18, using Definition 1.3 collapses the four cases into the single formula $f'(x) = \lim_{t \to x} \phi(t)$), and only at interior points is the derivative a genuinely stronger condition requiring the two one-sided derivatives to agree.

Let $f$ be a real-valued function on $[a, b]$.

(Case 2 : $x = a$.) Assume $f$ is differentiable at $a$. By Definition 1.17 this means that

exists. Writing (1.2) with $x = a$, we have $t \neq a$ for every $t \in (a, b)$, so

Now take $\lim_{t \to a^+}$ of both sides. Each of the two factors on the right has a right limit: the first has right limit $f'(a)$, and for the second, given any $\varepsilon > 0$, putting $\delta := \varepsilon$ gives, for all $t \in (a, a + \delta)$,

so $\lim_{t \to a^+} (t - a) = 0$. Hence by Exercise 1.14(ii) (the product rule for right limits),

The constant function $t \mapsto f(a)$ has right limit $f(a)$ at $a$ by Exercise 1.9, so from $f(t) = (f(t) - f(a)) + f(a)$ and Exercise 1.14(i) (the sum rule for right limits),

Thus $f$ is right continuous at $a$, and by Exercise 1.16(ii) this is the same as $f$ being continuous at $a$ in the sense of Definition 1.15.

(Case 3 : $x = b$.) This is symmetric. If $f$ is differentiable at $b$, then by Definition 1.17

exists. Write (1.2) with $x = b$ and take $\lim_{t \to b^-}$ of both sides. Given any $\varepsilon > 0$, putting $\delta := \varepsilon$ gives $|(t - b) - 0| = b - t \lt \varepsilon$ for all $t \in (b - \delta, b)$, so $\lim_{t \to b^-} (t - b) = 0$, and hence by Exercise 1.14(ii) (the product rule for left limits),

By Exercise 1.9 and Exercise 1.14(i), $\lim_{t \to b^-} f(t) = f(b)$, that is, $f$ is left continuous at $b$, which by Exercise 1.16(ii) is the same as $f$ being continuous at $b$ in the sense of Definition 1.15.

Together with Case 1 this proves Theorem 1.21 for every $x \in [a, b]$.

1. "Continuity at a point" is not determined by $f$ near that point. Definition 1.15 uses the local-sounding phrase "continuous at $c$", but its meaning is not determined by the behaviour of $f$ near $c$ alone: it also depends on which interval we regard $f$ as a function on. As we saw in Exercise 1.16(ii), at an endpoint the condition silently weakens to one-sided continuity. Concretely, consider

Regarded as a function on $[-1, 1]$, $H$ is not continuous at $0$ (take $\varepsilon = 1/2$; for any $\delta > 0$ and any $x \in (-\delta, 0)$ we have $|H(x) - H(0)| = 1$). But regarded as a function on $[0, 1]$, the range of $x$ demanded by Definition 1.15 is $(-\delta, \delta) \cap [0, 1] = [0, \delta) \cap [0, 1]$, on which $H \equiv 1$, so $H$ is continuous at $0$. The formula and the point are unchanged, yet merely changing the domain flips the truth value of "continuous at $0$". This is at odds with the local flavour the word "continuous" suggests.

2. It merely restates an existing definition (Definition 1.4) for one particular shape of domain. By Remark 1.2 the set $[a, b]$ is a metric space with the metric inherited from $\mathbb{R}$, and as shown in Exercise 1.16(iii), applying Definition 1.4 with $X = E = [a, b]$, $Y = \mathbb{R}$ produces literally the same statement as Definition 1.15. So Definition 1.15 is not a new notion but one instance of Definition 1.4 written out by hand, and it ought to be a theorem (or an observation) rather than a definition. The moment it is made a definition, one is committed to rewriting the definition for every new shape of domain. What about a half-open interval $[a, b)$? A union of intervals? A region in $\mathbb{R}^k$, or a set such as $\mathbb{Q}$? Definition 1.4 handles them all at once, whereas the style of Definition 1.15 requires deciding afresh, for each shape of domain, how the intersection with $(c - \delta, c + \delta)$ should be cut. And that intersection is precisely what Definition 1.4 already accomplishes with the single clause $x \in E$.

3. Theorems and proofs become case-ridden. The price of treating endpoints specially shows up immediately in the theorems. The proof of Theorem 1.21 splits into Cases 1, 2, 3, two of which are deferred to Exercise 1.22. The proofs of Theorem 1.25 and Theorem 1.30 skip the endpoints altogether with the phrase "due to the similarity of the proof, we only consider the case $x \in (a, b)$". Filling those gaps requires both versions of the algebra of limits, Theorem 1.12 and Exercise 1.14, and, on the derivative side, the one-sided derivatives of Exercise 1.19 as well. On an open interval, by contrast, every point is interior, so the single condition $\lim_{x \to c} f(x) = f(c)$ and a single argument suffice. Case splitting not only lengthens arguments; it creates places where errors can hide inside the cases that were omitted.

4. What should be done instead. Closed intervals cannot simply be abandoned: as in Remark 1.24, the existence of a maximum and a minimum requires the domain to be compact, and the same is true of Rolle's theorem and the mean-value theorem later on. The trouble is not the use of closed intervals but defining "continuity" anew for them. The desirable arrangement is this: define continuity once and for all by Definition 1.4 (for metric spaces with the induced metric), and demote Definition 1.15 to a theorem that computes that special case. Then the word "continuous" always means the same thing, while the definition itself makes visible that it is a property not of a bare function $f$ but of the data $f : E \to Y$, domain included. The weakening of the condition at an endpoint is then not a hidden exception but a consequence of the single rule "intersect with $E$". And notions that genuinely need interior points, such as the derivative, can be discussed on the interior $(a, b)$, with the endpoints handled separately under the name of one-sided derivatives (Exercise 1.19). The lecture note introduced Definition 1.15 in order to state Theorem 1.21 on $[a, b]$; that goal is met equally well in the way just described, without paying the three prices above.

Definition 1.28 defines an angle as the length of the corresponding arc, and $\cos \theta$, $\sin \theta$ as the coordinates of the point $P$ on the unit circle corresponding to that angle. Below we first state explicitly the geometric facts read off from that figure; after that only Definition 1.6, Theorem 1.12 and Definition 1.17 are used.

(The geometric facts used.) Put $A := (1, 0)$ and, for $0 \lt \theta \lt \pi/2$, let $P = P(\theta) := (\cos\theta, \sin\theta)$ be the point of the unit circle corresponding to the angle $\theta$.

(G1) The arc $\overset{\frown}{AP}$ has length $\theta$ (this is the definition of the angle in Definition 1.28 itself).

(G2) $P$ lies on the unit circle, so $|OP| = 1$, that is,

(G3) The shortest curve joining two points is the segment; in particular the length of a chord is at most the length of the arc.

(G4) If convex regions satisfy $K_1 \subset K_2$, then the perimeter of $K_1$ is at most the perimeter of $K_2$ (Archimedes' convexity principle).

(G5) Reflection in the $x$-axis preserves lengths and carries the arc of angle $\theta$ to the arc of angle $-\theta$, so $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$. Likewise a rotation about the origin preserves lengths and carries the point of angle $h$ to the point of angle $x + h$; in coordinates a rotation is

so applying it to $(u, v) = (\cos h, \sin h)$ gives the addition formulas

(The lecture note treats trigonometric identities as known, and we likewise use these as known facts coming from the geometry of Definition 1.28.)

Step 1: $\sin\theta \le \theta$ for $0 \lt \theta \lt \pi/2$. Put $Q := (\cos\theta, 0)$. The triangle $APQ$ has a right angle at $Q$, with $|PQ| = \sin\theta$ and hypotenuse $|PA|$. In a right triangle a leg is at most the hypotenuse, so $\sin\theta = |PQ| \le |PA|$, and $|PA| \le \theta$ by (G3) and (G1). Hence

Step 2: $1 - \cos\theta \le \theta^2 / 2$ for $0 \lt \theta \lt \pi/2$. By (G2),

and combining this with $|PA| \le \theta$ from Step 1 gives $2 - 2\cos\theta \le \theta^2$, that is,

Step 3: $\theta \le 2\tan(\theta/2)$, hence $u \le \tan u$ for $0 \lt u \lt \pi/4$. Let $0 \lt \theta \lt \pi/2$ and let $T$ be the intersection of the tangent line to the unit circle at $A$ with the tangent line at $P$ (they are not parallel, so they meet). A tangent is perpendicular to the radius at the point of tangency, so the triangle $OAT$ has a right angle at $A$; by symmetry $T$ lies on the bisector of the angle $AOP$, so $\angle AOT = \theta/2$. Since $|OA| = 1$,

Now let $K_1$ be the sector $OAP$ (the region bounded by the segment $OA$, the arc $\overset{\frown}{AP}$ and the segment $PO$) and $K_2$ the quadrilateral $OATP$. Both are convex, and $K_1 \subset K_2$ because the unit disc lies in the half-planes determined by the two tangent lines. Comparing perimeters by (G4) and (G1),

that is, $\theta \le 2\tan(\theta/2)$. Putting $u := \theta/2$, for $0 \lt u \lt \pi/4$,

(The last rearrangement used $\cos u > 0$ for $0 \lt u \lt \pi/4$; indeed Step 2 gives $1 - \cos u \le u^2/2 \lt \pi^2/32 \lt 1$.)

Step 4: $\displaystyle \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. Put $q(\theta) := \dfrac{\sin\theta}{\theta}$ for $0 \lt |\theta| \lt \pi/4$. By (G5) the function $\sin$ is odd, so $q$ is even, i.e. $q(-\theta) = q(\theta)$. For $0 \lt \theta \lt \pi/4$, Steps 1 and 3 together give

and hence, by Step 2,

Since $q$ and $\theta^2$ are both even, this inequality holds for every $\theta$ with $0 \lt |\theta| \lt \pi/4$. Now let $\varepsilon > 0$ be given and put

Then for all $\theta \in (-\delta, 0) \cup (0, \delta)$,

so $\lim_{\theta \to 0} \dfrac{\sin\theta}{\theta} = 1$ by Definition 1.6(i). In the same way, the inequalities of Step 2 and (G5) give

(use $|\cos\theta - 1| \le \theta^2/2$ with $\delta := \min\{\pi/4, \sqrt{2\varepsilon}\}$, and $|\sin\theta| \le |\theta|$ with $\delta := \min\{\pi/4, \varepsilon\}$, respectively).

Step 5: $\displaystyle \lim_{\theta \to 0} \frac{\cos\theta - 1}{\theta} = 0$. Multiplying by the conjugate and using (G2), for $0 \lt |\theta| \lt \pi/4$,

(the denominator is nonzero because $1 + \cos\theta \ge 1 > 0$). By Step 4 and Theorem 1.12 the limit of each factor on the right exists, so

Step 6: assembling the difference quotients. Fix $x \in \mathbb{R}$. Since $\cos$ and $\sin$ are defined on all of $\mathbb{R}$, we may take any open interval and regard $x$ as an interior point, so by Definition 1.17 (and Remark 1.18) the derivative is given by $\lim_{t \to x}$. For $t \neq x$ put $h := t - x$; the addition formulas of (G5) give

and

Here the condition $0 \lt |t - x| \lt \delta$ is the same as $0 \lt |h| \lt \delta$, so the limits of Steps 4 and 5 may be read directly as limits as $t \to x$, namely

Applying the limit of a constant function (Exercise 1.9) and Theorem 1.12(i), (ii) to the two identities above,

and

Since both limits exist, $\cos$ and $\sin$ are differentiable at every point of $\mathbb{R}$.

We keep the setting of the rigorous proof of Theorem 1.30: $f$ is a real-valued function on $[a, b]$, $g$ is a real-valued function on a set $I$ containing the range of $f$, $f$ is differentiable at $x \in (a, b)$, $g$ is differentiable at $f(x)$, and $f(x) \in (c, d) \subset I$ for some $c \lt d$. Moreover

What has to be shown is

The heart of the proof is that $E_{g'}$ also has a value at $s = f(x)$ and that this value is $0$; that is, $E_{g'}$ is continuous at $f(x)$. Thanks to this, nothing goes wrong even if there are $t$ with $f(t) = f(x)$.

Let $\varepsilon > 0$ be given.

Step 1 (the $\delta_1$ on the $E_{g'}$ side). Since $g$ is differentiable at $f(x)$, Definition 1.17 gives

Hence by Definition 1.6(i) there exists a $\delta_1 > 0$ such that $(f(x) - \delta_1, f(x) + \delta_1) \subset (c, d)$ and

On the other hand, at $s = f(x)$ we have $|E_{g'}(f(x))| = 0 \lt \varepsilon$ by definition. Combining the two, the estimate holds on the whole neighbourhood, not merely the punctured one:

Step 2 (the $\delta$ on the $f$ side). Since $f$ is differentiable at $x$, Theorem 1.21 shows that $f$ is continuous at $x$, and as $x \in (a, b)$, Definition 1.15 gives $\lim_{t \to x} f(t) = f(x)$. Feeding the $\delta_1 > 0$ of Step 1 into Definition 1.6(i) in place of $\varepsilon$, there exists a $\delta > 0$ such that

Shrinking $\delta$ if necessary, we may also assume $(x - \delta, x) \cup (x, x + \delta) \subset (a, b)$.

Step 3 (composition). Let $t \in (x - \delta, x) \cup (x, x + \delta)$. Then $s := f(t)$ is an element of the domain $I$ of $g$ and $|s - f(x)| \lt \delta_1$ by Step 2, so $(*)$ gives

Since $\varepsilon > 0$ was arbitrary, Definition 1.6(i) yields $\lim_{t \to x} E_{g'}(f(t)) = 0$, which is (1.5).

Note that Step 2 provided only $|f(t) - f(x)| \lt \delta_1$, not $f(t) \neq f(x)$. Indeed there may be $t$ arbitrarily close to $x$ with $f(t) = f(x)$. Had $(*)$ held only on the punctured neighbourhood $0 \lt |s - f(x)| \lt \delta_1$, Step 3 would have broken down at such $t$. The definition $E_{g'}(f(x)) := 0$ is exactly what fills that gap. This is how the rigorous proof avoids the flaw of the intuitive proof pointed out in Exercise 1.32.

The intuitive proof has flaws in three different places.

(a) One cannot divide by $f(t) - f(x)$. On the second line, numerator and denominator are multiplied by $f(t) - f(x)$ to write

But if $f(t) = f(x)$, the first factor on the right is $0/0$ and means nothing. And this is no exceptional situation. If $f$ is a constant function, for example, then $f(t) - f(x) = 0$ for every $t$ and the right-hand side is defined nowhere. For a more serious example, consider $f$ of Example 1.34(iv),

at $x = 0$. This $f$ is differentiable at $0$ with $f'(0) = 0$, yet putting $t_n := \dfrac{1}{n\pi}$ for $n \in \mathbb{N}$ gives $\sin(1/t_n) = \sin(n\pi) = 0$, hence $f(t_n) = 0 = f(0)$. Since $t_n \to 0$, every punctured neighbourhood of $0$ contains points with $f(t) = f(0)$. The limit in Definition 1.6(i) is a condition required at every point of a punctured neighbourhood, and the right-hand side above fails to be defined throughout any such neighbourhood. So the limit in question cannot even be written down, let alone evaluated.

(b) Splitting the limit of a product is not justified. On the third line, the limit of a product is split into the product of two limits. Theorem 1.12(ii) permits this only when the limit of each factor exists, and nothing so far has shown that the limit of the first factor exists. As (a) shows, that factor may not even be defined, so this step effectively assumes what is to be proved.

(c) One cannot replace $\lim_{t \to x}$ by $\lim_{f(t) \to f(x)}$. The symbol $\lim_{f(t) \to f(x)}$ on the fourth line is defined neither in Definition 1.3 nor in Definition 1.6. Read charitably, it is meant to be

The step then amounts to the substitution $s = f(t)$, turning a limit as $t \to x$ into a limit as $s \to f(x)$. For such a substitution (a limit of a composition) to be legitimate, two things are needed. One is that $f(t) \to f(x)$ as $t \to x$, which is true by Theorem 1.21. The other is that $f(t) \neq f(x)$ for $t \neq x$ near $x$ — precisely the condition that (a) shows may fail. A punctured limit says nothing about the value at $s = f(x)$, so if there are $t$ arbitrarily close to $x$ with $f(t) = f(x)$, the substitution has no ground to stand on.

The rigorous proof avoids all three flaws. There, nothing is ever divided by $f(t) - f(x)$. The identity

holds for all $s \in I$, including $s = f(x)$ (where both sides are $0$), and the difficulty with the substitution is resolved by the definition $E_{g'}(f(x)) := 0$, which makes $E_{g'}$ continuous at $f(x)$. The outcome is (1.5), proved in Exercise 1.31.

Since $p \in \mathbb{Q}$, we may write $p = m/n$ with $m \in \mathbb{Z}$ and $n \in \mathbb{N}$. For $x > 0$, $x^{1/n}$ denotes the unique positive $y$ with $y^n = x$ (we use its existence and uniqueness as known facts coming from the completeness of $\mathbb{R}$; uniqueness follows since $t \mapsto t^n$ is strictly increasing on $(0, \infty)$), and $x^{m/n} := \left( x^{1/n} \right)^m$. We also use, as algebraic facts on $(0, \infty)$, the laws of rational exponents $x^r x^s = x^{r+s}$ and $\left( x^{1/n} \right)^k = x^{k/n}$, and likewise that $x^p$ does not depend on the chosen representation of $p$.

Below $x > 0$ is a fixed point; it is an interior point of, say, $[x/2, 2x]$, so by Definition 1.17 (and Remark 1.18) the derivative is given by the two-sided limit $\lim_{t \to x}$. Logarithms and exponential functions have not been introduced yet and are not used.

Step 1: natural exponents. For $k \in \mathbb{N}$ put $u_k(x) := x^k$. Then $u_k$ is differentiable on $\mathbb{R}$ and $u_k'(x) = kx^{k-1}$. We argue by induction on $k$. For $k = 1$, Example 1.20 gives $u_1'(x) = 1 = 1 \cdot x^0$. Assuming the claim for $k$, we have $u_{k+1} = u_k u_1$, so Theorem 1.25(ii) gives

Moreover $u_0 \equiv 1$ is a constant function, so Example 1.20 gives $u_0'(x) = 0 = 0 \cdot x^{-1}$ for $x \neq 0$.

Step 2: negative integer exponents. Let $k \in \mathbb{N}$ and $x > 0$. Since $x^k \neq 0$, formula (1.3) of Theorem 1.25 together with Step 1 gives

Combined with Step 1, for every $m \in \mathbb{Z}$ and every $x > 0$,

Step 3: differentiating the $n$-th root. Fix $n \in \mathbb{N}$ and put $g(x) := x^{1/n}$ for $x > 0$. We use the algebraic identity: for all $y, y_0$,

(Expanding the right-hand side, neighbouring terms cancel and the left-hand side remains.)

(3a) Continuity of $g$. Fix $x_0 > 0$ and, for $x > 0$, put $y := g(x)$ and $y_0 := g(x_0)$. Then $y, y_0 > 0$ with $y^n = x$ and $y_0^n = x_0$, so (B) gives

(every term of the sum is positive and the $k = 0$ term is $y_0^{n-1}$). Hence

Given $\varepsilon > 0$, put $\delta := \min \left\{ \dfrac{x_0}{2}, \ \varepsilon \, y_0^{n-1} \right\} > 0$. If $|x - x_0| \lt \delta$, then $x > x_0 / 2 > 0$, so $g(x)$ is defined and $|g(x) - g(x_0)| \lt \varepsilon$. Thus $g$ is continuous at $x_0$, and in particular

(3b) Differentiability of $g$. Let $x > 0$ with $x \neq x_0$. Since $t \mapsto t^n$ is one-to-one on $(0, \infty)$, we have $y \neq y_0$, and (B) gives

(the denominator is nonzero because each of its terms is positive). As $x \to x_0$ we have $y = g(x) \to y_0$ by (3a), so repeated use of the limit of a constant function (Exercise 1.9) and of Theorem 1.12(ii) gives, for each $k$,

and repeated use of Theorem 1.12(i) gives

Hence Theorem 1.12(iii) yields

Since $x_0 > 0$ was arbitrary, for all $x > 0$,

that is, the stated formula holds in the case $p = 1/n$.

Step 4: general $p = m/n$. Let $x > 0$. We have $g(x) = x^{1/n} > 0$, the function $u_m(s) = s^m$ is differentiable on $(0, \infty)$ by Step 2, and $g$ is differentiable at $x$ by Step 3. Since $f(x) = x^{m/n} = \left( x^{1/n} \right)^m = u_m(g(x))$, Theorem 1.30 (the chain rule) together with (A) and (C) gives

Finally $\dfrac{m-n}{n} = \dfrac{m}{n} - 1 = p - 1$, so