Contents / Real Analysis / Differentiation

Chapter 4

Differentiation

The derivative from the difference quotient; differentiability implies continuity; the differentiation rules proved from the definition, with the chain rule via Caratheodory's formulation; Rolle, the Mean Value Theorem and monotonicity; Cauchy's MVT and L'Hopital's rule; Taylor's theorem with Lagrange, integral, Cauchy and Peano remainders; and the standard counterexamples of a discontinuous derivative and a smooth non-analytic function.

Introduction

The derivative is the first genuinely local construction in analysis. Continuity, proved in Continuity at a point, only asks that $f (x)$ stay near $f (a)$ ; differentiability asks something far stronger — that $f$ agree with a straight line to an error which is small compared with the distance travelled. That extra demand is what buys the whole of the calculus: a linear model accurate to first order can be added, multiplied, composed and inverted by rules, and once it exists at every point of an interval it constrains the function globally.

The chapter has four movements. The first builds the derivative from the difference quotient, extracts the linear-approximation formulation that makes every differentiation rule a one-line computation, and proves the rules. The second is the Mean Value Theorem, which is where local information becomes global: from " $f^{'} > 0$ at every point" — a statement about infinitely many separate instants — we shall deduce " $f$ is increasing", a statement about pairs of points arbitrarily far apart. Nothing in the definition of the derivative makes that deduction obvious, and the bridge is built out of the Extreme Value Theorem. The third movement iterates the idea: if one derivative gives a line, $n$ derivatives give a degree- $n$ polynomial, with an error term we can bound exactly. The fourth cashes Taylor's machinery in for L'Hôpital's rule.

Two counterexamples run through the chapter and are worth naming in advance, because each kills a plausible belief. The function $x^{2} sin (1/ x)$ is differentiable at every real number and yet its derivative is discontinuous at $0$ — so "differentiable" does not imply "continuously differentiable", and the derivative of a function need not be a function one can plot without lifting the pen. The function $e^{- 1/ x^{2}}$ is infinitely differentiable everywhere and has every derivative equal to zero at the origin — so its Taylor series there is identically $0$ and converges to the wrong function everywhere except one point. The first shows that the derivative operator loses regularity; the second shows that infinitely many derivatives still do not determine a function. Both are elementary, and both are the reason the theorems below are stated with the hypotheses they carry rather than shorter ones.

Throughout, $I$ denotes an interval with more than one point, and all functions are real-valued functions of a real variable.

4.1The Difference Quotient and the Derivative

The average rate of change of $f$ across $[a, a + h]$ is the slope of the chord joining the two endpoints of the graph. The derivative is what that slope becomes when the chord collapses to a point — provided it becomes anything at all.

Definition 4.1 (Derivative at a point). Let $f$ be defined on an open interval containing $a$ . The derivative of $f$ at $a$ is

f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h} = x \to a lim \frac{f ( x ) - f ( a )}{x - a},

whenever this limit exists and is finite, in which case $f$ is differentiable at $a$ . If $f$ is differentiable at every point of an interval $I$ , it is differentiable on $I$ , and $f^{'} : I \to R$ is its derivative function.

The two displayed limits are the same limit written in two variables, $x = a + h$ ; the second form is the more useful one in proofs, the first in computations. Both require $f$ to be defined on a neighbourhood of $a$ , because the limit must be approached from both sides. That is not a technicality — it is what makes the derivative a two-sided constraint.

Definition 4.2 (One-sided derivatives). The right-hand and left-hand derivatives of $f$ at $a$ are

f_{+}^{'} (a) = h \to 0^{+} lim \frac{f ( a + h ) - f ( a )}{h}, f_{-}^{'} (a) = h \to 0^{-} lim \frac{f ( a + h ) - f ( a )}{h},

when these limits exist. At an endpoint of a closed interval, "differentiable" means the appropriate one-sided derivative exists.

By the theorem Two-sided limits from one-sided limits, $f$ is differentiable at an interior point $a$ exactly when $f_{+}^{'} (a)$ and $f_{-}^{'} (a)$ both exist and are equal. A corner is precisely the failure of that equality.

The difference quotient is awkward to manipulate: it is undefined at $h = 0$ , which is the only value we care about. The following reformulation removes the quotient entirely, and with it most of the epsilon-chasing in the proofs that follow.

Proposition 4.3 (Linear approximation and Carathéodory's formulation). Let $f$ be defined near $a$ and let $L \in R$ . The following are equivalent.

$f$ is differentiable at $a$ with $f^{'} (a) = L$ .
$f (a + h) = f (a) + L h + ε (h) h$ where $ε (h) \to 0$ as $h \to 0$ (equivalently, the error is $o (h)$ ).
There is a function $φ$ , defined near $a$ and continuous at $a$ , with $φ (a) = L$ and

f (x) - f (a) = φ (x) (x - a) for all x near a .

Proof. $(1) \Rightarrow (2)$ . Define $ε (h) = \frac{f ( a + h ) - f ( a )}{h} - L$ for $h \neq = 0$ and $ε (0) = 0$ . Rearranging gives the displayed identity, and $ε (h) \to 0$ is exactly the statement $f^{'} (a) = L$ .

$(2) \Rightarrow (3)$ . Set $φ (x) = L + ε (x - a)$ for $x \neq = a$ and $φ (a) = L$ . Then $φ (x) (x - a) = L (x - a) + ε (x - a) (x - a) = f (x) - f (a)$ by $(2)$ , and $φ (x) \to L = φ (a)$ as $x \to a$ , so $φ$ is continuous at $a$ .

$(3) \Rightarrow (1)$ . For $x \neq = a$ divide the identity by $x - a$ to get $\frac{f ( x ) - f ( a )}{x - a} = φ (x)$ . Letting $x \to a$ and using continuity of $φ$ at $a$ , the quotient tends to $φ (a) = L$ .∎

Form $(2)$ is the one to carry in the head: differentiable means well approximated by a line, with error negligible against $h$ . Form $(3)$ , due to Carathéodory, is the one to carry into proofs, because it has no division in it and therefore survives composition without a case analysis. We shall use it for the chain rule, where the naive argument breaks.

Example 4.4 (A corner: the absolute value at the origin). Show that $f (x) = ∣ x ∣$ is not differentiable at $0$ , and identify its one-sided derivatives there.

Solution.

For $h > 0$ , $\frac{∣0 + h ∣ - ∣0∣}{h} = \frac{h}{h} = 1$ , so $f_{+}^{'} (0) = 1$ .
For $h < 0$ , $\frac{∣ h ∣}{h} = \frac{- h}{h} = - 1$ , so $f_{-}^{'} (0) = - 1$ .
The one-sided limits exist but differ, so by the theorem Two-sided limits from one-sided limits the two-sided limit does not exist and $f^{'} (0)$ is undefined.
Note what did not fail: $∣ x ∣$ is continuous at $0$ , and it is differentiable at every other point. Non-differentiability here is a single-point defect caused by the graph having two distinct tangent directions rather than one.

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Remark. Definition Derivative at a point demands a finite limit. Some texts write $g^{'} (0) = + \infty$ for $x^{1/3}$ ; we do not, because every theorem in this chapter uses $f^{'} (a)$ as a real number in an arithmetic identity, and admitting $\pm \infty$ would break them all.

Exercises for §4.1 →

4.2Differentiability, Continuity, and Higher Derivatives

Differentiability is strictly stronger than continuity. The proof is three lines and the converse is false, and the gap between the two is much larger than the single counterexample suggests.

Theorem 4.5 (Differentiability implies continuity). If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ .

Proof. By the proposition Linear approximation and Carathéodory's formulation, there is $φ$ continuous at $a$ with $f (x) - f (a) = φ (x) (x - a)$ near $a$ . Both factors on the right are continuous at $a$ , and the second vanishes there, so by the theorem Limit laws

x \to a lim (f (x) - f (a)) = φ (a) \cdot 0 = 0.

Hence $f (x) \to f (a)$ , which is continuity at $a$ .∎

The converse fails, and the example A corner: the absolute value at the origin already shows it at one point. It fails much more dramatically than that: Weierstrass exhibited a function continuous on all of $R$ and differentiable at no point, so the implication cannot be reversed even approximately. Continuity is a constraint on values; differentiability is a constraint on how the values are approached.

Definition 4.6 (Higher derivatives and smoothness classes). If $f^{'}$ is itself differentiable on $I$ , its derivative is $f^{''}$ , and inductively $f^{(n)} = (f^{(n - 1)})^{'}$ , with $f^{(0)} = f$ . Write $f \in C^{n} (I)$ if $f^{(n)}$ exists and is continuous on $I$ , and $f \in C^{\infty} (I)$ — smooth — if $f \in C^{n} (I)$ for every $n$ .

The distinction between " $f$ is $n$ times differentiable" and " $f \in C^{n}$ " looks pedantic and is not. The second is strictly stronger, and the function that separates them is the first of the chapter's two standing counterexamples.

Proposition 4.7 (A differentiable function with a discontinuous derivative). Define

f (x) = {x^{2} sin (1/ x), 0, x \neq = 0, x = 0.

Then $f$ is differentiable at every real number, with $f^{'} (0) = 0$ , but $f^{'}$ is not continuous at $0$ . Hence $f$ is differentiable everywhere yet $f \in / C^{1} (R)$ .

Proof. For $x \neq = 0$ the function is a product and composite of functions differentiable there, so by the theorem Differentiation rules and the theorem Chain rule below,

f^{'} (x) = 2 x sin (1/ x) - cos (1/ x) .

At the origin we must use the definition, since the formula above is not available:

\frac{f ( h ) - f ( 0 )}{h} = \frac{h ^{2} sin ( 1/ h )}{h} = h sin (1/ h),

and $∣ h sin (1/ h) ∣ \leq ∣ h ∣ \to 0$ , so by the Squeeze Theorem the limit exists and $f^{'} (0) = 0$ .

Now let $x \to 0$ along the sequence $x_{n} = 1/ (2 πn)$ . Then $sin (1/ x_{n}) = 0$ and $cos (1/ x_{n}) = 1$ , so $f^{'} (x_{n}) = - 1$ for every $n$ , while $f^{'} (0) = 0$ . By the theorem Sequential characterization of continuity, $f^{'}$ is discontinuous at $0$ . The term $2 x sin (1/ x)$ does tend to $0$ ; it is $cos (1/ x)$ , which oscillates without settling, that destroys continuity.∎

Intuition. Read the counterexample as a statement about scales. Near the origin the graph of $x^{2} sin (1/ x)$ is squeezed between the parabolas $\pm x^{2}$ , so it is flattened into the origin fast enough for the tangent to exist and be horizontal. But it wiggles infinitely often on the way in, and each wiggle, however small in height, is steep — a wiggle of height $x^{2}$ across a horizontal distance of order $x^{2}$ has slope of order $1$ . So the height of the oscillation dies (giving differentiability) while its slope does not (destroying continuity of the derivative).

Pitfall. " $f$ is differentiable, so I may differentiate $f^{'}$ " is a non-sequitur, and " $f^{'}$ exists near $a$ , so $f^{'}$ is continuous at $a$ " is false, by the proposition A differentiable function with a discontinuous derivative. Whenever a theorem below asks for $f \in C^{1}$ or for continuity of $f^{(n)}$ , that hypothesis is doing real work and cannot be weakened to mere existence.

Example 4.8 (Matching one-sided derivatives to make a $C^{1}$ join). Find $a, b$ so that $f (x) = x^{2}$ for $x \leq 1$ and $f (x) = a x + b$ for $x > 1$ is differentiable at $1$ .

Solution.

Continuity at $1$ is necessary by the theorem Differentiability implies continuity: $lim_{x \to 1^{-}} f = 1$ and $lim_{x \to 1^{+}} f = a + b$ , so $a + b = 1$ .
The left derivative is $f_{-}^{'} (1) = 2 \cdot 1 = 2$ ; the right derivative of a line is its slope, $a$ .
Equality of one-sided derivatives forces $a = 2$ , hence $b = 1 - a = - 1$ .
So $f (x) = 2 x - 1$ for $x > 1$ , the tangent line to $x^{2}$ at $x = 1$ — which is the only possible answer, since a line agreeing with $f$ in value and slope at $1$ is the tangent.

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Exercises for §4.2 →

4.3The Rules of Differentiation and the Chain Rule

Every differentiation rule is a statement that the linear approximation of a combination is the corresponding combination of the linear approximations. Proving them from the definition is short once the algebra is arranged correctly, and the arrangement is the content.

Theorem 4.9 (Algebra of derivatives). Let $f$ and $g$ be differentiable at $a$ and $c \in R$ . Then $f + g$ , $c f$ and $f g$ are differentiable at $a$ , and so is $f / g$ provided $g (a) \neq = 0$ , with

(f + g)^{'} (a) = f^{'} (a) + g^{'} (a), (c f)^{'} (a) = c f^{'} (a),

(f g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a), (\frac{f}{g})^{'} (a) = \frac{f ^{'} ( a ) g ( a ) - f ( a ) g ^{'} ( a )}{g ( a ) ^{2}} .

Proof. Sum and scalar multiple. The difference quotient of $f + g$ is the sum of the two difference quotients, and the limit of a sum is the sum of the limits by the theorem Limit laws; the same argument gives $(c f)^{'} = c f^{'}$ .

Product. Add and subtract the cross term $f (a + h) g (a)$ in the numerator:

\frac{f ( a + h ) g ( a + h ) - f ( a ) g ( a )}{h} = f (a + h) \frac{g ( a + h ) - g ( a )}{h} + g (a) \frac{f ( a + h ) - f ( a )}{h} .

As $h \to 0$ the two quotients tend to $g^{'} (a)$ and $f^{'} (a)$ , and $f (a + h) \to f (a)$ because $f$ is continuous at $a$ by the theorem Differentiability implies continuity — this is where that theorem earns its place. The limit is $f (a) g^{'} (a) + g (a) f^{'} (a)$ .

Reciprocal. Since $g$ is continuous at $a$ and $g (a) \neq = 0$ , there is a neighbourhood of $a$ on which $g \neq = 0$ , so the quotient is defined there. Then

\frac{1}{h} (\frac{1}{g ( a + h )} - \frac{1}{g ( a )}) = \frac{- 1}{g ( a + h ) g ( a )} \cdot \frac{g ( a + h ) - g ( a )}{h} ⟶ \frac{- g ^{'} ( a )}{g ( a ) ^{2}} .

Quotient. Apply the product rule to $f \cdot (1/ g)$ :

(\frac{f}{g})^{'} (a) = \frac{f ^{'} ( a )}{g ( a )} - \frac{f ( a ) g ^{'} ( a )}{g ( a ) ^{2}} = \frac{f ^{'} ( a ) g ( a ) - f ( a ) g ^{'} ( a )}{g ( a ) ^{2}} .

∎

Theorem 4.10 (Chain rule). Suppose $g$ is differentiable at $a$ and $f$ is differentiable at $g (a)$ . Then $f \circ g$ is differentiable at $a$ and

(f \circ g)^{'} (a) = f^{'} (g (a)) g^{'} (a) .

Proof. Write $b = g (a)$ . By the proposition Linear approximation and Carathéodory's formulation applied to $f$ at $b$ , there is $φ$ continuous at $b$ with $φ (b) = f^{'} (b)$ and

f (y) - f (b) = φ (y) (y - b)

for all $y$ near $b$ . Substituting $y = g (x)$ , which is legitimate for $x$ near $a$ because $g$ is continuous at $a$ and so keeps $g (x)$ near $b$ ,

f (g (x)) - f (g (a)) = φ (g (x)) (g (x) - g (a)) .

Divide by $x - a$ for $x \neq = a$ :

\frac{f ( g ( x )) - f ( g ( a ))}{x - a} = φ (g (x)) \cdot \frac{g ( x ) - g ( a )}{x - a} .

As $x \to a$ , the second factor tends to $g^{'} (a)$ . For the first, $g$ is continuous at $a$ and $φ$ is continuous at $b = g (a)$ , so $φ \circ g$ is continuous at $a$ and $φ (g (x)) \to φ (b) = f^{'} (b)$ . The product of the limits is $f^{'} (g (a)) g^{'} (a)$ .∎

Remark. The naive proof multiplies and divides by $g (x) - g (a)$ and fails, because $g (x) - g (a)$ can vanish at points arbitrarily close to $a$ — take $g \equiv 0$ , or $g (x) = x^{2} sin (1/ x)$ , whose zeros accumulate at the origin. Carathéodory's formulation avoids the division altogether, which is exactly why it was worth proving.

Proposition 4.11 (Derivative of an inverse function). Let $f$ be continuous and strictly monotone on an interval $I$ , differentiable at $a \in I$ with $f^{'} (a) \neq = 0$ , and let $b = f (a)$ . Then $f^{- 1}$ is differentiable at $b$ and

(f^{- 1})^{'} (b) = \frac{1}{f ^{'} ( a )} = \frac{1}{f ^{'} ( f ^{- 1} ( b ) )} .

Proof. By the theorem Intermediate Value Theorem, a continuous strictly monotone $f$ maps $I$ onto an interval $J$ and $f^{- 1} : J \to I$ is continuous and strictly monotone. Let $y \to b$ in $J$ with $y \neq = b$ , and put $x = f^{- 1} (y)$ , so $x \neq = a$ and $x \to a$ by continuity of $f^{- 1}$ . Then

\frac{f ^{- 1} ( y ) - f ^{- 1} ( b )}{y - b} = \frac{x - a}{f ( x ) - f ( a )} = (\frac{f ( x ) - f ( a )}{x - a})^{- 1} .

The inner quotient tends to $f^{'} (a) \neq = 0$ , so by the theorem Limit laws the reciprocal tends to $1/ f^{'} (a)$ .∎

Example 4.12 (The derivative of $arctan$ ). Use the proposition Derivative of an inverse function to differentiate $arctan$ .

Solution.

Take $f = tan$ on $(- π /2, π /2)$ , which is continuous, strictly increasing, and has $f^{'} = sec^{2} = 1 + tan^{2} > 0$ throughout.
With $b = tan a$ , the proposition gives $(arctan)^{'} (b) = 1/ (1 + tan^{2} a)$ .
Since $tan a = b$ , this is $\frac{1}{1 + b ^{2}}$ . Renaming the variable, $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$ .
Sanity check: the answer is positive, as it must be for an increasing function, and it tends to $0$ as $∣ x ∣ \to \infty$ , matching the horizontal asymptotes at $\pm π /2$ .

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Exercises for §4.3 →

4.4From Local to Global: Interior Extrema

Everything so far is local bookkeeping: the derivative at $a$ is computed from values of $f$ arbitrarily close to $a$ , and no statement yet relates $f$ at two distant points. The bridge is a single observation about maxima, and it is the reason the Extreme Value Theorem appears in a chapter about derivatives at all.

Theorem 4.13 (Interior extremum theorem). Let $f$ be defined on $(a, b)$ and let $c \in (a, b)$ be a point at which $f$ attains a local maximum or a local minimum. If $f$ is differentiable at $c$ , then $f^{'} (c) = 0$ .

Proof. Suppose $c$ is a local maximum, so $f (c + h) \leq f (c)$ for all sufficiently small $∣ h ∣$ .

For such $h > 0$ the quotient $\frac{f ( c + h ) - f ( c )}{h}$ has a non-positive numerator and a positive denominator, so it is $\leq 0$ ; letting $h \to 0^{+}$ preserves the weak inequality, giving $f^{'} (c) \leq 0$ .

For such $h < 0$ the numerator is still $\leq 0$ but the denominator is negative, so the quotient is $\geq 0$ ; letting $h \to 0^{-}$ gives $f^{'} (c) \geq 0$ .

Both limits equal $f^{'} (c)$ , since $f$ is differentiable at $c$ , so $f^{'} (c) \leq 0 \leq f^{'} (c)$ and therefore $f^{'} (c) = 0$ . For a local minimum apply the result to $- f$ .∎

Three hypotheses are each indispensable, and dropping any one produces a counterexample rather than a harder proof. Interior: $f (x) = x$ on $[0, 1]$ attains both extrema, at endpoints, with $f^{'} \equiv 1$ . Differentiable: $f (x) = ∣ x ∣$ has a minimum at $0$ where $f^{'}$ does not exist. Extremum: $f (x) = x^{3}$ has $f^{'} (0) = 0$ without an extremum there — so the converse is false, and $f^{'} (c) = 0$ is necessary but not sufficient.

Call $c$ a critical point of $f$ if $f^{'} (c) = 0$ or $f^{'} (c)$ fails to exist. In that language the theorem Interior extremum theorem says: every interior local extremum is a critical point.

That last sentence is the whole optimisation method: to find the extrema of a continuous $f$ on $[a, b]$ , the Extreme Value Theorem guarantees they exist, and the theorem Interior extremum theorem confines them to the critical points together with the two endpoints — a search over a typically finite set instead of an interval.

Example 4.14 (Optimisation on a closed interval). Find the absolute maximum and minimum of $f (x) = x^{3} - 3 x$ on $[- 3, 2]$ .

Solution.

$f$ is a polynomial, hence continuous on a closed bounded interval, so by the Extreme Value Theorem both extrema are attained.
$f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$ , so the critical points are $x = \pm 1$ , both interior.
Evaluate at the critical points and the endpoints: $f (- 3) = - 27 + 9 = - 18$ , $f (- 1) = - 1 + 3 = 2$ , $f (1) = 1 - 3 = - 2$ , $f (2) = 8 - 6 = 2$ .
The maximum value is $2$ , attained twice — at the interior point $- 1$ and at the endpoint $2$ — and the minimum is $- 18$ at $x = - 3$ .
Note that the minimum sits at an endpoint where $f^{'} (- 3) = 24 \neq = 0$ ; searching only for zeros of $f^{'}$ would have missed it entirely.

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The theorem Interior extremum theorem is the last purely local result in the chapter. Everything from here — Rolle, the Mean Value Theorem, monotonicity, Taylor's theorem, L'Hôpital's rule — is a consequence of combining it with the Extreme Value Theorem, and every one of those results has the same shape: a hypothesis about $f^{'}$ at every point of an interval, a conclusion about $f$ at the interval's ends.

Exercises for §4.4 →

4.5The Mean Value Theorem: Statement and Geometry

Theorem 4.15 (Mean Value Theorem). Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there exists $c \in (a, b)$ with

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}, equivalently f (b) - f (a) = f^{'} (c) (b - a) .

The proof is deferred to the next section, where Rolle's theorem supplies it in two lines. First, what the statement says.

Geometrically: the chord joining $(a, f (a))$ to $(b, f (b))$ has slope equal to the average rate of change, and the theorem asserts that somewhere strictly inside the interval the tangent is parallel to that chord. Physically: if a car covers $180$ kilometres in two hours, at some instant its speedometer read exactly $90$ km/h. The average is not merely approached — it is attained, exactly, at some interior instant.

Figure 4.16. The chord of

x

over

[0, 4]

has slope

\frac{1}{2}

, and the tangent at

c = 1

is parallel to it.

Two features of the statement deserve emphasis because they are routinely misread. The theorem is existential: it asserts that some $c$ exists and gives no procedure for finding it, no formula for it, and no claim that it is unique — for $f (x) = sin x$ on $[0, 2 π]$ there are two such points. And the conclusion places $c$ in the open interval, which is what makes the theorem usable: a $c$ allowed to sit at an endpoint would say nothing, since $f^{'}$ need not even exist there.

The asymmetry in the hypotheses — continuity on the closed interval, differentiability only on the open one — is deliberate and is what lets the theorem apply to $f (x) = x$ on $[0, 4]$ , whose derivative blows up at the left endpoint. In the figure above, $f^{'} (x) = 1/ (2 x)$ is unbounded near $0$ , yet the conclusion holds with $c$ given by $1/ (2 c) = 1/2$ , that is $c = 1$ .

Example 4.17 (Locating $c$ explicitly). Find every $c$ guaranteed by the Mean Value Theorem for $f (x) = x^{3}$ on $[- 1, 2]$ .

Solution.

The chord slope is $\frac{f ( 2 ) - f ( - 1 )}{2 - ( - 1 )} = \frac{8 - ( - 1 )}{3} = 3$ .
Solve $f^{'} (c) = 3 c^{2} = 3$ , giving $c = \pm 1$ .
Only $c = 1$ lies in the open interval $(- 1, 2)$ ; the value $c = - 1$ is the endpoint and is excluded.
So the guaranteed point is unique here, $c = 1$ — but uniqueness is an accident of this $f$ , not part of the theorem.

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Exercises for §4.5 →

4.6Rolle's Theorem, and the Proof of the Mean Value Theorem

Rolle's theorem is the Mean Value Theorem in the special case of a horizontal chord. It looks weaker and is in fact equivalent, because tilting the picture is an algebraic operation that costs nothing.

Theorem 4.18 (Rolle's theorem). Let $f$ be continuous on $[a, b]$ , differentiable on $(a, b)$ , and suppose $f (a) = f (b)$ . Then there exists $c \in (a, b)$ with $f^{'} (c) = 0$ .

Proof. Since $f$ is continuous on the closed bounded interval $[a, b]$ , the Extreme Value Theorem says $f$ attains a maximum value $M$ and a minimum value $m$ on $[a, b]$ .

Case 1: $M = m$ . Then $f$ is constant on $[a, b]$ , so $f^{'} (c) = 0$ for every $c \in (a, b)$ , and the interval is non-empty since $a < b$ .

Case 2: $M > m$ . Since $f (a) = f (b)$ , that common value cannot equal both $M$ and $m$ ; say $M \neq = f (a)$ (otherwise run the argument with $m$ ). Let $c \in [a, b]$ be a point with $f (c) = M$ . Then $c \neq = a$ and $c \neq = b$ , because $f (a) = f (b) \neq = M$ , so $c$ lies in the open interval $(a, b)$ . There $f$ has a local — indeed global — maximum and is differentiable by hypothesis, so the theorem Interior extremum theorem gives $f^{'} (c) = 0$ .∎

Everything in that proof is inherited: the existence of the extremum comes from compactness via the Extreme Value Theorem, and the vanishing of the derivative comes from the theorem Interior extremum theorem. The only new idea is the bookkeeping that forces the extremum to be interior, and it is the hypothesis $f (a) = f (b)$ that supplies it.

Proof. Proof of the theorem Mean Value Theorem. Subtract the chord. Define, for $x \in [a, b]$ ,

g (x) = f (x) - f (a) - \frac{f ( b ) - f ( a )}{b - a} (x - a) .

Then $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , because it is $f$ minus a polynomial. Moreover $g (a) = 0$ and

g (b) = f (b) - f (a) - \frac{f ( b ) - f ( a )}{b - a} (b - a) = 0,

so $g (a) = g (b)$ . By the theorem Rolle's theorem there is $c \in (a, b)$ with $g^{'} (c) = 0$ , that is

f^{'} (c) - \frac{f ( b ) - f ( a )}{b - a} = 0,

which is the claim.∎

Intuition. The function $g$ measures the vertical gap between the graph and its own chord. That gap is zero at both ends, so it must turn around somewhere in between, and at the turning point its derivative vanishes — which says the graph and the chord are momentarily changing at the same rate. Rolle's theorem is thus not a special case that got lucky; it is the Mean Value Theorem viewed in a coordinate system where the chord is the horizontal axis.

Example 4.19 (Rolle's theorem counts roots). Show that $p (x) = x^{3} + 3 x + 1$ has exactly one real root.

Solution.

At least one. $p (- 1) = - 3 < 0$ and $p (0) = 1 > 0$ , so the Intermediate Value Theorem gives a root in $(- 1, 0)$ .
At most one. Suppose $p$ had two distinct roots $r < s$ . Then $p$ is continuous on $[r, s]$ , differentiable on $(r, s)$ , and $p (r) = p (s) = 0$ , so the theorem Rolle's theorem supplies $c \in (r, s)$ with $p^{'} (c) = 0$ .
But $p^{'} (x) = 3 x^{2} + 3 \geq 3 > 0$ for every real $x$ , so no such $c$ exists — a contradiction.
Hence exactly one real root. The argument used no root-finding at all: the derivative's sign alone limited the number of roots.

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Exercises for §4.6 →

4.7Consequences: Monotonicity, Bounds, and Inequalities

The Mean Value Theorem converts pointwise information about $f^{'}$ into statements about $f$ on whole intervals. Every result in this section is the same one-line deduction applied to a different hypothesis on the sign or size of $f^{'}$ .

Corollary 4.20 (Zero derivative means constant). If $f$ is continuous on an interval $I$ and $f^{'} (x) = 0$ for every interior point $x$ , then $f$ is constant on $I$ . Consequently, if $f^{'} = g^{'}$ throughout $I$ , then $f - g$ is constant.

Proof. Let $x < y$ in $I$ . Then $f$ is continuous on $[x, y]$ and differentiable on $(x, y)$ , so the theorem Mean Value Theorem gives $c \in (x, y)$ with

f (y) - f (x) = f^{'} (c) (y - x) = 0.

Hence $f (y) = f (x)$ for every pair, so $f$ is constant. For the second claim apply the first to $f - g$ , whose derivative is identically $0$ by the theorem Algebra of derivatives.∎

Pitfall. The hypothesis " $I$ is an interval" cannot be dropped. On the set $(- \infty, 0) \cup (0, \infty)$ the function $sign (x)$ has derivative $0$ everywhere and is not constant. The Mean Value Theorem needs $[x, y]$ to lie inside the domain, which fails when the domain is disconnected — and this is precisely why " $+ C$ " in an antiderivative means one constant per interval.

Theorem 4.21 (Monotonicity from the sign of the derivative). Let $f$ be continuous on an interval $I$ and differentiable at every interior point.

If $f^{'} \geq 0$ throughout, $f$ is non-decreasing on $I$ ; if $f^{'} \leq 0$ throughout, $f$ is non-increasing.
If $f^{'} > 0$ throughout, $f$ is strictly increasing; if $f^{'} < 0$ throughout, $f$ is strictly decreasing.

Proof. Let $x < y$ in $I$ . By the theorem Mean Value Theorem there is $c \in (x, y)$ with

f (y) - f (x) = f^{'} (c) (y - x),

and $y - x > 0$ . So the sign of $f (y) - f (x)$ is exactly the sign of $f^{'} (c)$ .

If $f^{'} \geq 0$ everywhere then $f (y) - f (x) \geq 0$ , giving (1); if $f^{'} > 0$ everywhere then $f (y) - f (x) > 0$ strictly, giving (2). The decreasing cases follow by applying these to $- f$ .∎

The converse of part (2) is false in one direction only, and the example is worth remembering: $f (x) = x^{3}$ is strictly increasing on $R$ yet $f^{'} (0) = 0$ . So strict monotonicity forces $f^{'} \geq 0$ but not $f^{'} > 0$ . What is true is that $f^{'} \geq 0$ with $f^{'}$ vanishing only at isolated points still gives strict increase, since the proof above yields a strict inequality as soon as $f^{'}$ is non-zero somewhere in $(x, y)$ .

Corollary 4.22 (Derivative bounds give Lipschitz bounds). If $f$ is differentiable on an interval $I$ with $∣ f^{'} (x) ∣ \leq M$ for all $x \in I$ , then

∣ f (y) - f (x) ∣ \leq M ∣ y - x ∣ for all x, y \in I .

Proof. For $x \neq = y$ the theorem Mean Value Theorem on the interval with endpoints $x$ and $y$ gives $c$ strictly between them with $f (y) - f (x) = f^{'} (c) (y - x)$ . Take absolute values and use $∣ f^{'} (c) ∣ \leq M$ . The case $x = y$ is trivial.∎

This is how the Mean Value Theorem is used in practice: not to find $c$ , but to throw it away after bounding $f^{'} (c)$ . The unknown $c$ is harmless precisely because the bound holds at every point.

Example 4.23 (Proving an inequality by differentiation). Show that $ln (1 + x) \leq x$ for every $x > - 1$ , with equality only at $x = 0$ .

Solution.

Set $g (x) = x - ln (1 + x)$ on $(- 1, \infty)$ , so the claim is $g \geq 0$ .
$g^{'} (x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x}$ , whose denominator is positive on the domain. So $g^{'} < 0$ on $(- 1, 0)$ and $g^{'} > 0$ on $(0, \infty)$ .
By the theorem Monotonicity from the sign of the derivative, $g$ is strictly decreasing on $(- 1, 0]$ and strictly increasing on $[0, \infty)$ , so its minimum over the whole domain is at $x = 0$ .
$g (0) = 0 - ln 1 = 0$ , hence $g (x) \geq 0$ everywhere with equality only at $0$ , which is the claim.

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Theorem 4.24 (Cauchy Mean Value Theorem). Let $f$ and $g$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there exists $c \in (a, b)$ with

(f (b) - f (a)) g^{'} (c) = (g (b) - g (a)) f^{'} (c) .

If in addition $g^{'}$ never vanishes on $(a, b)$ , then $g (b) \neq = g (a)$ and the conclusion may be divided:

\frac{f ( b ) - f ( a )}{g ( b ) - g ( a )} = \frac{f ^{'} ( c )}{g ^{'} ( c )} .

The statement is recorded here because the next sections need it; its proof is given in the section Proving Cauchy's Mean Value Theorem, where it belongs to the discussion of L'Hôpital's rule. Taking $g (x) = x$ recovers the theorem Mean Value Theorem, so it is a genuine generalisation.

Exercises for §4.7 →

4.8When the Hypotheses Are Really Needed

The Mean Value Theorem carries three hypotheses: continuity on the closed interval, differentiability on the open interval, and real-valuedness. Each is sharp, and the counterexamples are small.

Example 4.25 (Dropping continuity at an endpoint). Let $f (x) = x$ for $x \in [0, 1)$ and $f (1) = 0$ . Show the conclusion of the Mean Value Theorem fails on $[0, 1]$ .

Solution.

$f$ is differentiable on $(0, 1)$ with $f^{'} \equiv 1$ , so the differentiability hypothesis holds.
But $f$ is not continuous at $1$ : $lim_{x \to 1^{-}} f (x) = 1 \neq = 0 = f (1)$ .
The chord slope is $\frac{f ( 1 ) - f ( 0 )}{1 - 0} = \frac{0 - 0}{1} = 0$ , yet $f^{'} (c) = 1$ for every $c \in (0, 1)$ .
No $c$ works. A single endpoint discontinuity is enough: it lets the chord slope be chosen freely without any constraint on $f^{'}$ inside.

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Example 4.26 (Dropping differentiability at one interior point). Show the conclusion fails for $f (x) = ∣ x ∣$ on $[- 1, 2]$ .

Solution.

$f$ is continuous on $[- 1, 2]$ and differentiable except at the single point $0$ .
The chord slope is $\frac{∣2∣ - ∣ - 1∣}{2 - ( - 1 )} = \frac{1}{3}$ .
But $f^{'} (x) = - 1$ for $x < 0$ and $f^{'} (x) = + 1$ for $x > 0$ ; the value $1/3$ is never attained.
So one bad interior point destroys the conclusion. The derivative here takes only the values $\pm 1$ — it skips $1/3$ — which is possible only because $f^{'}$ is undefined at $0$ ; a derivative that exists on a whole interval cannot skip values, by Darboux's theorem.

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Remark. Notice what is not required: $f^{'}$ need not be continuous, need not be bounded, and need not exist at either endpoint. The theorem applies to $f (x) = x^{2} sin (1/ x)$ from the proposition A differentiable function with a discontinuous derivative and to $f (x) = x$ on $[0, 1]$ without modification. Weakening hypotheses to exactly what the proof consumes is what makes a theorem reusable, and the proof consumed only the Extreme Value Theorem, which needs continuity on a closed bounded interval, plus differentiability where the extremum can land.

Exercises for §4.8 →

4.9Taylor's Theorem with the Lagrange Remainder

The Mean Value Theorem says $f (b) = f (a) + f^{'} (c) (b - a)$ : the function is a constant plus an exactly-corrected linear term. Taylor's theorem is the same statement with a degree- $n$ polynomial in place of the constant, and it is proved by the same device — build an auxiliary function with enough vanishing and apply Rolle repeatedly.

Definition 4.27 (Taylor polynomial). Let $f$ be $n$ times differentiable at $a$ . The $n$ -th Taylor polynomial of $f$ at $a$ is

P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( a )}{k !} (x - a)^{k},

and the remainder is $R_{n} (x) = f (x) - P_{n} (x)$ . When $a = 0$ this is also called the Maclaurin polynomial.

$P_{n}$ is the unique polynomial of degree at most $n$ that matches $f$ in value and in the first $n$ derivatives at $a$ : differentiating the sum $k$ times and setting $x = a$ kills every term but one, leaving $P_{n}^{(k)} (a) = f^{(k)} (a)$ for $0 \leq k \leq n$ . Everything interesting is in $R_{n}$ .

Theorem 4.28 (Taylor's theorem, Lagrange form of the remainder). Let $f$ be $n + 1$ times differentiable on an open interval $I$ containing $a$ , with $f^{(n)}$ continuous on the closed interval between $a$ and $x$ . Then for each $x \in I$ with $x \neq = a$ there is a point $ξ$ strictly between $a$ and $x$ such that

f (x) = k = 0 \sum n \frac{f ^{(k)} ( a )}{k !} (x - a)^{k} + \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} (x - a)^{n + 1} .

For $n = 0$ this is precisely the theorem Mean Value Theorem.

Proof. Fix $x \neq = a$ and choose the number $M$ so that

f (x) = P_{n} (x) + M (x - a)^{n + 1};

this is possible, and defines $M$ uniquely, because $(x - a)^{n + 1} \neq = 0$ . The claim is that $M = f^{(n + 1)} (ξ) / (n + 1)!$ for some $ξ$ between $a$ and $x$ .

Define, for $t$ in the closed interval between $a$ and $x$ ,

g (t) = f (t) - P_{n} (t) - M (t - a)^{n + 1} .

Then $g$ is $n + 1$ times differentiable there. Two families of zeros are available:

$g (a) = 0$ , and more: since $P_{n}^{(k)} (a) = f^{(k)} (a)$ for $k = 0, \dots, n$ , and every derivative of $(t - a)^{n + 1}$ up to order $n$ vanishes at $t = a$ , we get $g^{(k)} (a) = 0$ for $k = 0, 1, \dots, n$ .
$g (x) = 0$ by the choice of $M$ .

Now apply the theorem Rolle's theorem repeatedly. Since $g (a) = g (x) = 0$ , there is $ξ_{1}$ strictly between $a$ and $x$ with $g^{'} (ξ_{1}) = 0$ . Since $g^{'} (a) = 0 = g^{'} (ξ_{1})$ , there is $ξ_{2}$ strictly between $a$ and $ξ_{1}$ with $g^{''} (ξ_{2}) = 0$ . Continuing, after $n + 1$ applications we obtain $ξ = ξ_{n + 1}$ strictly between $a$ and $x$ with

g^{(n + 1)} (ξ) = 0.

Compute that derivative. $P_{n}$ has degree at most $n$ , so $P_{n}^{(n + 1)} \equiv 0$ , and $\frac{d ^{n + 1}}{d t ^{n + 1}} (t - a)^{n + 1} = (n + 1)!$ . Hence

0 = g^{(n + 1)} (ξ) = f^{(n + 1)} (ξ) - M (n + 1)!,

so $M = f^{(n + 1)} (ξ) / (n + 1)!$ , which is the assertion.∎

The remainder term is the $(n + 1)$ -st term of the series with one change: the derivative is evaluated not at $a$ but at an unknown interior point $ξ$ . That single substitution is what makes the formula an identity rather than an approximation, and — exactly as with the Mean Value Theorem — the practical use is to bound $∣ f^{(n + 1)} ∣$ on the interval and discard $ξ$ .

Example 4.29 (The remainder for the exponential). Write $e^{x}$ with its degree- $3$ Maclaurin polynomial and Lagrange remainder, and bound the error for $∣ x ∣ \leq 1$ .

Solution.

All derivatives of $f (x) = e^{x}$ are $e^{x}$ , so $f^{(k)} (0) = 1$ and $P_{3} (x) = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}$ .
The theorem Taylor's theorem, Lagrange form of the remainder gives $R_{3} (x) = \frac{e ^{ξ}}{24} x^{4}$ for some $ξ$ between $0$ and $x$ .
For $∣ x ∣ \leq 1$ we have $ξ \leq 1$ , so $e^{ξ} \leq e < 3$ and $∣ R_{3} (x) ∣ \leq \frac{3}{24} = 0.125$ .
At $x = 1$ : $P_{3} (1) = 2.6 \overline{6}$ against $e = 2.71828 \dots$ , an actual error of $0.0516$ — inside the bound, as it must be, and about half of it, since the bound used the worst-case $e^{ξ}$ .

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Exercises for §4.9 →

4.10Other Forms of the Remainder

Lagrange's form is the sharpest tool when $f^{(n + 1)}$ can be bounded. When it cannot, or when the remainder must be manipulated rather than estimated, two other forms are better.

Theorem 4.30 (Integral form of the remainder). If $f^{(n + 1)}$ is continuous on the closed interval between $a$ and $x$ , then

R_{n} (x) = f (x) - P_{n} (x) = \frac{1}{n !} \int_{a}^{x} (x - t)^{n} f^{(n + 1)} (t) d t .

Proof. Induct on $n$ . For $n = 0$ the claim is $f (x) - f (a) = \int_{a}^{x} f^{'} (t) d t$ , which is the theorem Fundamental Theorem of Calculus, Part 2, applicable since $f^{'}$ is continuous.

Assume the formula for $n - 1$ . Integrate the $n - 1$ case by parts, taking $u = f^{(n)} (t)$ and $d v = \frac{( x - t ) ^{n - 1}}{( n - 1 )!} d t$ , so that $v = - \frac{( x - t ) ^{n}}{n !}$ :

\frac{1}{( n - 1 )!} \int_{a}^{x} (x - t)^{n - 1} f^{(n)} (t) d t = [- \frac{( x - t ) ^{n}}{n !} f^{(n)} (t)]_{a}^{x} + \frac{1}{n !} \int_{a}^{x} (x - t)^{n} f^{(n + 1)} (t) d t .

The boundary term vanishes at $t = x$ and equals $\frac{( x - a ) ^{n}}{n !} f^{(n)} (a)$ at $t = a$ . So

R_{n - 1} (x) = \frac{f ^{(n)} ( a )}{n !} (x - a)^{n} + \frac{1}{n !} \int_{a}^{x} (x - t)^{n} f^{(n + 1)} (t) d t,

and since $R_{n - 1} (x) - \frac{f ^{(n)} ( a )}{n !} (x - a)^{n} = R_{n} (x)$ by the definition Taylor polynomial, the induction closes.∎

The integral form requires more of $f$ — continuity of $f^{(n + 1)}$ , so that the integral exists — and gives more in return: no unknown point, and a remainder that can be estimated, integrated again, or differentiated under the integral sign.

Corollary 4.31 (Lagrange and Cauchy forms from the integral form). Suppose $f^{(n + 1)}$ is continuous between $a$ and $x$ . Then there exist points $ξ, η$ strictly between $a$ and $x$ with

R_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} (x - a)^{n + 1} (Lagrange),

R_{n} (x) = \frac{f ^{(n + 1)} ( η )}{n !} (x - η)^{n} (x - a) (Cauchy) .

Proof. Take $x > a$ ; the case $x < a$ is identical with the orientation reversed.

Cauchy form. Apply the mean value theorem for integrals to $\frac{1}{n !} \int_{a}^{x} (x - t)^{n} f^{(n + 1)} (t) d t$ , treating the whole integrand as a continuous function: there is $η \in (a, x)$ with the integral equal to $\frac{( x - η ) ^{n} f ^{(n + 1)} ( η )}{n !} (x - a)$ .

Lagrange form. Instead apply the weighted mean value theorem for integrals with weight $w (t) = (x - t)^{n} \geq 0$ , which does not change sign: there is $ξ \in (a, x)$ with

\frac{1}{n !} \int_{a}^{x} (x - t)^{n} f^{(n + 1)} (t) d t = \frac{f ^{(n + 1)} ( ξ )}{n !} \int_{a}^{x} (x - t)^{n} d t = \frac{f ^{(n + 1)} ( ξ )}{n !} \cdot \frac{( x - a ) ^{n + 1}}{n + 1},

which is the Lagrange expression. The two forms differ only in which factor was pulled out of the integral.∎

A third form asks nothing about $f^{(n + 1)}$ at all, and says nothing quantitative in exchange — only that the remainder is of smaller order than the last term kept.

Proposition 4.32 (Peano form of the remainder). If $f$ is $n$ times differentiable at $a$ , then

f (x) = P_{n} (x) + o ((x - a)^{n}) as x \to a,

that is, $R_{n} (x) / (x - a)^{n} \to 0$ .

Proof. For $n = 1$ this is condition (2) of the proposition Linear approximation and Carathéodory's formulation. For general $n$ , both $R_{n} (x)$ and $(x - a)^{n}$ tend to $0$ as $x \to a$ , and $R_{n}^{(k)} (a) = 0$ for $k = 0, \dots, n$ by construction of $P_{n}$ . Apply the theorem L'Hôpital's rule, the $0/0$ form $n - 1$ times:

x \to a lim \frac{R _{n} ( x )}{( x - a ) ^{n}} = x \to a lim \frac{R _{n}^{(n - 1)} ( x )}{n ! ( x - a ) /1},

where at each stage numerator and denominator both vanish at $a$ , so the hypotheses are met. The last limit is

\frac{1}{n !} x \to a lim \frac{R _{n}^{(n - 1)} ( x ) - R _{n}^{(n - 1)} ( a )}{x - a} = \frac{R _{n}^{(n)} ( a )}{n !} = 0,

using only that $R_{n}^{(n - 1)}$ is differentiable at $a$ — which is all the hypothesis gives — and that $R_{n}^{(n)} (a) = 0$ .∎

Remark. The three forms answer three different questions. Lagrange: how big can the error be? Integral: what exactly is the error, as a formula I can operate on? Peano: is the error negligible against the last term I kept? Peano's is the one used to compute limits, because it needs only $n$ derivatives at the single point $a$ — the cheapest hypothesis in the chapter.

Exercises for §4.10 →

4.11Taylor Series and When They Represent the Function

Letting $n \to \infty$ in the definition Taylor polynomial produces a power series. Whether that series converges, and whether it converges to $f$ , are two separate questions, and both can fail.

Definition 4.33 (Taylor series; analytic functions). For $f$ smooth near $a$ , the Taylor series of $f$ at $a$ is the formal power series

k = 0 \sum \infty \frac{f ^{(k)} ( a )}{k !} (x - a)^{k} .

$f$ is analytic at $a$ if this series converges to $f (x)$ on some open interval around $a$ , and analytic on a set if it is analytic at each of its points.

Theorem 4.34 (Remainder criterion for series representation). Let $f$ be smooth on an interval $I$ containing $a$ . Then the Taylor series of $f$ at $a$ converges to $f (x)$ for a given $x \in I$ if and only if $R_{n} (x) \to 0$ as $n \to \infty$ . In particular, if there are constants $M, r > 0$ with $∣ f^{(n)} (t) ∣ \leq M r^{n}$ for all $n$ and all $t$ between $a$ and $x$ , then the series converges to $f (x)$ .

Proof. The $n$ -th partial sum of the series is exactly $P_{n} (x)$ , so by the definition of $R_{n}$ ,

f (x) - k = 0 \sum n \frac{f ^{(k)} ( a )}{k !} (x - a)^{k} = ∣ R_{n} (x) ∣.

The left side tends to $0$ precisely when the series converges to $f (x)$ , and the right side tends to $0$ precisely when $R_{n} (x) \to 0$ ; the two statements are therefore the same statement.

For the sufficient condition, the theorem Taylor's theorem, Lagrange form of the remainder gives

∣ R_{n} (x) ∣ = \frac{∣ f ^{(n + 1)} ( ξ ) ∣}{( n + 1 )!} ∣ x - a ∣^{n + 1} \leq M \frac{( r ∣ x - a ∣ ) ^{n + 1}}{( n + 1 )!} \cdot r^{0} .

For fixed $s = r ∣ x - a ∣$ the sequence $s^{n + 1} / (n + 1)!$ tends to $0$ , since it is the general term of the convergent series for $e^{s}$ and the theorem Term test forces the terms of a convergent series to vanish. Hence $R_{n} (x) \to 0$ .∎

Example 4.35 (The exponential series converges everywhere). Prove that $e^{x} = \sum_{k \geq 0} x^{k} / k!$ for every real $x$ .

Solution.

Fix $x$ and let $A = max (∣ x ∣, 1)$ . Every derivative of $e^{t}$ is $e^{t}$ , and for $t$ between $0$ and $x$ we have $e^{t} \leq e^{A}$ .
So $∣ f^{(n)} (t) ∣ \leq e^{A} = e^{A} \cdot 1^{n}$ : the hypothesis of the theorem Remainder criterion for series representation holds with $M = e^{A}$ and $r = 1$ .
Therefore $R_{n} (x) \to 0$ and the series converges to $e^{x}$ for that $x$ . Since $x$ was arbitrary, this holds on all of $R$ .
The same argument with $∣ f^{(n)} ∣ \leq 1$ handles $sin$ and $cos$ , which is why those three series are the ones quoted without further comment.

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Convergence of the series is not the issue; convergence to $f$ is. The following function is the standard demonstration that smoothness is not enough.

Proposition 4.36 (A smooth function that is not analytic). Define

f (x) = {e^{- 1/ x^{2}}, 0, x \neq = 0, x = 0.

Then $f \in C^{\infty} (R)$ and $f^{(n)} (0) = 0$ for every $n \geq 0$ . Its Taylor series at $0$ is identically zero and therefore converges to $f$ only at $x = 0$ ; $f$ is not analytic at $0$ .

Proof. For $x \neq = 0$ , repeated use of the theorem Chain rule and the theorem Algebra of derivatives shows by induction that

f^{(n)} (x) = p_{n} (1/ x) e^{- 1/ x^{2}}

for some polynomial $p_{n}$ : the base case is $p_{0} = 1$ , and differentiating $p_{n} (1/ x) e^{- 1/ x^{2}}$ produces $(- x^{- 2} p_{n}^{'} (1/ x) + 2 x^{- 3} p_{n} (1/ x)) e^{- 1/ x^{2}}$ , again of that shape.

It remains to show $f^{(n)} (0) = 0$ for all $n$ , by induction. Suppose $f^{(n)} (0) = 0$ . Then

\frac{f ^{(n)} ( h ) - f ^{(n)} ( 0 )}{h} = \frac{p _{n} ( 1/ h ) e ^{- 1/ h^{2}}}{h} = q (1/ h) e^{- 1/ h^{2}}

where $q (u) = u p_{n} (u)$ is a polynomial. Put $u = 1/∣ h ∣$ , so $u \to \infty$ as $h \to 0$ , and the modulus of the expression is at most $∣ q ∣ (u) e^{- u^{2}}$ once $∣ q ∣$ denotes the polynomial with the absolute values of the coefficients. For any fixed degree $d$ , $u^{d} e^{- u^{2}} \to 0$ as $u \to \infty$ , since $e^{u^{2}} \geq u^{2 m} / m!$ for every $m$ by the series for the exponential, and choosing $2 m > d$ makes $u^{d} e^{- u^{2}} \leq m! u^{d - 2 m} \to 0$ . Hence the difference quotient tends to $0$ and $f^{(n + 1)} (0) = 0$ . Continuity of $f^{(n + 1)}$ at $0$ follows from the same estimate applied to $p_{n + 1} (1/ x) e^{- 1/ x^{2}}$ .

So all derivatives at $0$ vanish, the Taylor series at $0$ is $\sum 0 \cdot x^{k} = 0$ , and it converges (to $0$ ) for every $x$ — but $f (x) > 0$ for every $x \neq = 0$ , so it represents $f$ nowhere except at the origin.∎

Figure 4.37. Taylor polynomials of

sin

0

: each added term extends the interval of visible agreement, and each eventually diverges.

Intuition. A real-analytic function is rigid: its values on any tiny interval determine it everywhere it is analytic, because they determine every derivative at a point and hence the whole series. A merely smooth function is floppy: $e^{- 1/ x^{2}}$ is identically $0$ to infinite order at the origin and yet is strictly positive one step to the right. That floppiness is a feature, not a defect — it is exactly what makes bump functions possible, and bump functions are the reason smooth partitions of unity exist while analytic ones do not.

Exercises for §4.11 →

4.12Applications: Approximation and Error Bounds

The practical content of Taylor's theorem is a two-step routine: write down $P_{n}$ , then bound $∣ f^{(n + 1)} ∣$ on the interval of interest and read off the error. The second step is where the theorem does work no numerical experiment can do — it bounds the error without knowing the true value.

Figure 4.38. The actual error

∣ e^{x} - P_{2} (x) ∣

[0, 1]

stays under the Lagrange bound

e ∣ x ∣^{3} /6

, which is what Taylor's theorem guarantees.

Example 4.39 (How many terms for four-decimal accuracy?). Find the smallest $n$ for which $P_{n}$ of $e^{x}$ at $0$ approximates $e^{0.5}$ with error below $1 0^{- 4}$ .

Solution.

By the theorem Taylor's theorem, Lagrange form of the remainder, $∣ R_{n} (0.5) ∣ = \frac{e ^{ξ}}{( n + 1 )!} (0.5)^{n + 1}$ with $0 < ξ < 0.5$ , so $e^{ξ} < e^{0.5} < 1.65$ .
Require $\frac{1.65 ( 0.5 ) ^{n + 1}}{( n + 1 )!} < 1 0^{- 4}$ .
Test values: $n = 4$ gives $\frac{1.65 \cdot 0.03125}{120} = 4.3 \times 1 0^{- 4}$ , too big; $n = 5$ gives $\frac{1.65 \cdot 0.015625}{720} = 3.6 \times 1 0^{- 5}$ , which passes.
So $n = 5$ suffices. Checking, $P_{5} (0.5) = 1.6487 \overline{2}$ against $e^{0.5} = 1.6487212 \dots$ , an error near $1 0^{- 6}$ — the bound was conservative by an order of magnitude, as bounds built on worst-case derivatives usually are.

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Exercises for §4.12 →

4.13L'Hôpital's Rule: the $0/0$ Form

A quotient whose numerator and denominator both tend to $0$ has no determined limit: $\frac{2 x}{x} \to 2$ , $\frac{x ^{2}}{x} \to 0$ and $\frac{x}{x ^{2}} \to \infty$ all have the form $0/0$ . The symbol $0/0$ records which limit laws fail, not a value.

A limit is called an indeterminate form when substituting the limiting behaviour of its parts yields one of

\frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, \infty - \infty, 0^{0}, 1^{\infty}, \infty^{0},

that is, an expression whose value is not determined by the limits of the parts alone.

The point of the list is its complement. $\frac{1}{0 ^{+}} = + \infty$ , $0^{\infty} = 0$ and $\infty + \infty = \infty$ are not indeterminate: those limits are determined, and applying L'Hôpital's rule to them is an error, not merely a waste.

Theorem 4.40 (L'Hôpital's rule, the $0/0$ form). Let $f, g$ be differentiable on an open interval $(a, b)$ with $g^{'} (x) \neq = 0$ there, and suppose

x \to a^{+} lim f (x) = x \to a^{+} lim g (x) = 0.

If $x \to a^{+} lim \frac{f ^{'} ( x )}{g ^{'} ( x )} = L$ exists, finite or infinite, then

x \to a^{+} lim \frac{f ( x )}{g ( x )} = L .

The same holds for left-hand limits, two-sided limits, and limits at $\pm \infty$ .

Proof. Extend $f$ and $g$ to $[a, b)$ by setting $f (a) = g (a) = 0$ . The hypothesis on the limits makes both extensions continuous at $a$ , and they remain differentiable on $(a, b)$ .

Fix $x \in (a, b)$ . The functions $f, g$ are continuous on $[a, x]$ and differentiable on $(a, x)$ , and $g^{'}$ does not vanish there, so the theorem Cauchy Mean Value Theorem applies and produces a point $c_{x} \in (a, x)$ with

\frac{f ( x )}{g ( x )} = \frac{f ( x ) - f ( a )}{g ( x ) - g ( a )} = \frac{f ^{'} ( c _{x} )}{g ^{'} ( c _{x} )} .

The first equality uses $f (a) = g (a) = 0$ ; the denominator $g (x) = g (x) - g (a)$ is non-zero by the second part of that theorem.

Now let $x \to a^{+}$ . Since $a < c_{x} < x$ , the Squeeze Theorem gives $c_{x} \to a^{+}$ . Therefore $\frac{f ^{'} ( c _{x} )}{g ^{'} ( c _{x} )} \to L$ by hypothesis, and by the identity above $\frac{f ( x )}{g ( x )} \to L$ as well.∎

The proof explains the hypotheses exactly. Both functions must vanish in the limit, or the first equality is false; $g^{'}$ must not vanish, or Cauchy's theorem cannot be divided; and the limit of $f^{'} / g^{'}$ must exist, because the rule transfers a known limit and does not create one.

Example 4.41 (A first application, and why one pass is not always enough). Evaluate $x \to 0 lim \frac{e ^{x} - 1 - x}{x ^{2}}$ .

Solution.

Numerator and denominator both tend to $0$ , and $g^{'} (x) = 2 x \neq = 0$ for $x \neq = 0$ , so the theorem L'Hôpital's rule, the $0/0$ form applies.
Differentiating: $lim_{x \to 0} \frac{e ^{x} - 1}{2 x}$ , again of the form $0/0$ .
Apply the rule a second time: $lim_{x \to 0} \frac{e ^{x}}{2} = \frac{1}{2}$ .
Since the last limit exists, the rule transfers backwards through both steps, and the original limit is $\frac{1}{2}$ . Checking with the proposition Peano form of the remainder: $e^{x} - 1 - x = \frac{x ^{2}}{2} + o (x^{2})$ , giving $\frac{1}{2}$ immediately.

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Exercises for §4.13 →

4.14The $\infty/\infty$ Form and the Other Indeterminates

Theorem 4.42 (L'Hôpital's rule, the $\infty/\infty$ form). Let $f, g$ be differentiable on $(a, b)$ with $g^{'} \neq = 0$ there, and suppose $∣ g (x) ∣ \to \infty$ as $x \to a^{+}$ . If $lim_{x \to a^{+}} f^{'} (x) / g^{'} (x) = L$ exists, finite or infinite, then

x \to a^{+} lim \frac{f ( x )}{g ( x )} = L .

Notably, nothing need be assumed about $f$ beyond differentiability.

Proof. Take $L$ finite; the infinite case follows by applying the finite one to $g / f$ . Let $ε > 0$ . Choose $y \in (a, b)$ so close to $a$ that

\frac{f ^{'} ( t )}{g ^{'} ( t )} - L < \frac{ε}{2} for all t \in (a, y) .

For $x \in (a, y)$ the theorem Cauchy Mean Value Theorem on $[x, y]$ gives $c \in (x, y) \subset (a, y)$ with

\frac{f ( x ) - f ( y )}{g ( x ) - g ( y )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}, so \frac{f ( x ) - f ( y )}{g ( x ) - g ( y )} - L < \frac{ε}{2} .

Now write the quotient we actually want in terms of that one:

\frac{f ( x )}{g ( x )} = \frac{f ( x ) - f ( y )}{g ( x ) - g ( y )} \cdot \to 1 \frac{g ( x ) - g ( y )}{g ( x )} + \frac{f ( y )}{g ( x )} .

Here $y$ is now fixed and $∣ g (x) ∣ \to \infty$ as $x \to a^{+}$ , so the marked factor tends to $1$ and the last term tends to $0$ . Hence for $x$ sufficiently close to $a$ the right-hand side differs from $\frac{f ( x ) - f ( y )}{g ( x ) - g ( y )}$ by less than $ε /2$ , and therefore from $L$ by less than $ε$ . As $ε$ was arbitrary, the limit is $L$ .∎

The device in that proof — freeze $y$ , let $x$ run — is the reason the $\infty/\infty$ form needs a separate argument rather than a substitution trick, and it is also why no hypothesis on $f$ is required: the term $f (y) / g (x)$ is killed by the growth of $g$ alone.

The remaining indeterminate forms are handled by rewriting them as quotients. A product of the form $0 \cdot \infty$ becomes $\frac{f}{1/ g}$ or $\frac{g}{1/ f}$ , and the two rewrites are not equally convenient — choose the one whose derivative is simpler, since the other frequently produces something worse than the original. A difference of the form $\infty - \infty$ is combined over a common denominator, or has its dominant term factored out, until it reads as $0/0$ or $\infty/\infty$ . The three exponential forms $1^{\infty}$ , $0^{0}$ and $\infty^{0}$ are reduced by logarithms: set $y = f^{g}$ , so that $ln y = g ln f$ is a $0 \cdot \infty$ form, evaluate $lim ln y = Λ$ , and conclude $lim y = e^{Λ}$ by continuity of the exponential.

For example, $(1 + \frac{3}{x})^{x}$ as $x \to \infty$ is $1^{\infty}$ ; its logarithm is $\frac{l n ( 1 + 3/ x )}{1/ x}$ , a $0/0$ form whose derivative quotient is $\frac{3}{1 + 3/ x} \to 3$ , so the original limit is $e^{3}$ .

Example 4.43 (Polynomials lose to exponentials). Show that $x \to \infty lim \frac{x ^{n}}{e ^{x}} = 0$ for every fixed positive integer $n$ .

Solution.

The form is $\infty/\infty$ and $g^{'} (x) = e^{x}$ never vanishes, so the theorem L'Hôpital's rule, the $\infty/\infty$ form applies.
One application lowers the power: $lim \frac{n x ^{n - 1}}{e ^{x}}$ , still $\infty/\infty$ .
After $n$ applications the numerator is the constant $n!$ and the limit is $lim \frac{n !}{e ^{x}} = 0$ , which exists.
Transferring back through the $n$ steps gives $0$ . This is the precise sense in which "exponential beats polynomial", and iterating once more shows $e^{x}$ likewise beats $x^{n}$ for every $n$ simultaneously.

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Exercises for §4.14 →

4.15Proving Cauchy's Mean Value Theorem

Both forms of L'Hôpital's rule rest on one generalisation of the Mean Value Theorem, stated earlier as the theorem Cauchy Mean Value Theorem. Its proof is the debt this chapter still owes, and it is settled in a few lines by the same trick used for the Mean Value Theorem itself: build an auxiliary function that vanishes at both endpoints, and call Rolle.

Proof. Proof of the theorem Cauchy Mean Value Theorem. Define, for $t \in [a, b]$ ,

h (t) = (f (b) - f (a)) g (t) - (g (b) - g (a)) f (t) .

Then $h$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , being a linear combination of $f$ and $g$ with constant coefficients. Evaluate at the endpoints:

h (a) = f (b) g (a) - f (a) g (a) - g (b) f (a) + g (a) f (a) = f (b) g (a) - g (b) f (a),

h (b) = f (b) g (b) - f (a) g (b) - g (b) f (b) + g (a) f (b) = g (a) f (b) - f (a) g (b) .

These are equal, so $h (a) = h (b)$ and the theorem Rolle's theorem supplies $c \in (a, b)$ with $h^{'} (c) = 0$ , that is

(f (b) - f (a)) g^{'} (c) = (g (b) - g (a)) f^{'} (c),

which is the first conclusion.

For the second, suppose $g^{'}$ never vanishes on $(a, b)$ . If $g (b) = g (a)$ then the theorem Rolle's theorem applied to $g$ alone would give a point where $g^{'} = 0$ , a contradiction; so $g (b) \neq = g (a)$ , and both $g^{'} (c)$ and $g (b) - g (a)$ are non-zero, permitting the division.∎

Intuition. Read the conclusion parametrically. The pair $(g (t), f (t))$ traces a curve in the plane as $t$ runs over $[a, b]$ ; the chord from $(g (a), f (a))$ to $(g (b), f (b))$ has slope $\frac{f ( b ) - f ( a )}{g ( b ) - g ( a )}$ , while the tangent vector to the curve at parameter $c$ is $(g^{'} (c), f^{'} (c))$ , of slope $\frac{f ^{'} ( c )}{g ^{'} ( c )}$ . Cauchy's theorem says the tangent is parallel to the chord somewhere — the same picture as the ordinary Mean Value Theorem, drawn for a parametrised curve rather than a graph. Taking $g (t) = t$ parametrises the graph of $f$ and recovers the original.

Example 4.44 (Cauchy's theorem used directly). For $f (x) = x^{2}$ and $g (x) = x^{3}$ on $[1, 2]$ , find the point $c$ of the theorem Cauchy Mean Value Theorem.

Solution.

$\frac{f ( 2 ) - f ( 1 )}{g ( 2 ) - g ( 1 )} = \frac{4 - 1}{8 - 1} = \frac{3}{7}$ .
$\frac{f ^{'} ( c )}{g ^{'} ( c )} = \frac{2 c}{3 c ^{2}} = \frac{2}{3 c}$ for $c \neq = 0$ .
Solve $\frac{2}{3 c} = \frac{3}{7}$ , giving $9 c = 14$ and $c = \frac{14}{9} \approx 1.556$ .
This lies in $(1, 2)$ , as promised. Note it is not the point produced by the ordinary Mean Value Theorem for either function separately — for $f$ that would be $c = 3/2$ .

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Exercises for §4.15 →

4.16Where L'Hôpital's Rule Goes Wrong

The rule is applied wrongly more often than any other result in elementary analysis, and every failure is one of four kinds.

Pitfall. Checking the form is not optional. Applying the rule to $lim_{x \to 0} \frac{c o s x}{x ^{2}}$ , which is $\frac{1}{0 ^{+}} = + \infty$ and hence determinate, gives $lim \frac{- s i n x}{2 x} = - \frac{1}{2}$ — a confident, wrong answer. Verify that numerator and denominator both tend to $0$ , or that the denominator tends to $\pm \infty$ , before differentiating.

Example 4.45 (The converse direction fails). Show that $x \to \infty lim \frac{x + sin x}{x}$ exists but that the quotient of derivatives has no limit.

Solution.

Directly: $\frac{x + s i n x}{x} = 1 + \frac{s i n x}{x}$ , and $\frac{s i n x}{x} \leq \frac{1}{x} \to 0$ , so the limit is $1$ .
The derivatives give $\frac{1 + c o s x}{1}$ , which oscillates between $0$ and $2$ forever and has no limit.
There is no contradiction: the theorem L'Hôpital's rule, the $\infty/\infty$ form concludes something if $lim f^{'} / g^{'}$ exists, and says nothing when it does not.
So "L'Hôpital fails" never means "the limit does not exist" — it means the tool was uninformative and another method is needed.

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Pitfall. It is not the quotient rule. The rule replaces $\frac{f}{g}$ by $\frac{f ^{'}}{g ^{'}}$ — the derivatives taken separately — which is emphatically not $(\frac{f}{g})^{'}$ . Writing $\frac{f ^{'} g - f g ^{'}}{g ^{2}}$ at this point is a different, and useless, computation.

Pitfall. The hypothesis $g^{'} \neq = 0$ is real. Both proofs divide by $g^{'} (c)$ for a point $c$ we do not control, so $g^{'}$ must be non-vanishing on a whole one-sided neighbourhood, not merely at points where it is convenient. A denominator like $g (x) = x + sin x cos x$ near $0$ , whose derivative vanishes on a sequence approaching the limit point, is outside the theorem's scope even though the form is $0/0$ .

Exercises for §4.16 →

Common Mistakes

Assuming a derivative is continuous. By the proposition A differentiable function with a discontinuous derivative, $x^{2} sin (1/ x)$ is differentiable everywhere with a derivative that is discontinuous at $0$ . "Differentiable" and "continuously differentiable" are different hypotheses and theorems distinguish them.
Reversing the theorem Differentiability implies continuity. Continuity gives nothing: $∣ x ∣$ is continuous and not differentiable at $0$ , and Weierstrass's function is continuous nowhere differentiable.
Treating $f^{'} (c) = 0$ as proof of an extremum. It is necessary at an interior extremum, not sufficient — $f (x) = x^{3}$ at $0$ . And extrema at endpoints or at non-differentiable points are not found this way at all.
Forgetting the endpoints when optimising. The Extreme Value Theorem guarantees the extrema exist somewhere in $[a, b]$ ; the theorem Interior extremum theorem only narrows down where the interior ones can be.
Solving for $c$ in the Mean Value Theorem when the problem does not ask. The theorem's power is that $c$ can be bounded and discarded, as in the corollary Derivative bounds give Lipschitz bounds. Only exercises explicitly ask for $c$ .
Applying the corollary Zero derivative means constant across a gap. A zero derivative forces constancy on each interval of the domain separately; $sign (x)$ is the counterexample on a disconnected domain.
Claiming a Taylor series always represents its function. By the proposition A smooth function that is not analytic, $e^{- 1/ x^{2}}$ has an identically zero Taylor series at $0$ . Convergence of the series and convergence to $f$ must be checked separately, via the theorem Remainder criterion for series representation.
Bounding the remainder with $f^{(n + 1)} (a)$ instead of $sup ∣ f^{(n + 1)} ∣$ . The Lagrange remainder is evaluated at an unknown $ξ$ between $a$ and $x$ , so the bound must hold over the entire interval, not just at the centre.
Applying L'Hôpital's rule to a determinate form. $\frac{1}{0}$ , $\frac{0}{5}$ and $\frac{\infty}{3}$ are not indeterminate, and differentiating them produces confident nonsense.
Concluding that a limit fails to exist because $f^{'} / g^{'}$ has no limit. The rule is one-directional; $\frac{x + s i n x}{x}$ has limit $1$ while $\frac{1 + c o s x}{1}$ has none.
Differentiating the quotient instead of the numerator and denominator separately. L'Hôpital's rule is not the quotient rule, and confusing the two changes the problem.
Iterating L'Hôpital without rechecking the form. Each pass needs the hypotheses verified again; once the form becomes determinate, the next differentiation is invalid and the answer silently wrong.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

4.1The Difference Quotient and the Derivative

4.2Differentiability, Continuity, and Higher Derivatives

4.3The Rules of Differentiation and the Chain Rule

4.4From Local to Global: Interior Extrema

4.5The Mean Value Theorem: Statement and Geometry

4.6Rolle's Theorem, and the Proof of the Mean Value Theorem

4.7Consequences: Monotonicity, Bounds, and Inequalities

4.8When the Hypotheses Are Really Needed

4.9Taylor's Theorem with the Lagrange Remainder

4.10Other Forms of the Remainder

4.11Taylor Series and When They Represent the Function

4.12Applications: Approximation and Error Bounds

4.13L'Hôpital's Rule: the 0/0 Form

4.14The ∞/∞ Form and the Other Indeterminates

4.15Proving Cauchy's Mean Value Theorem

4.16Where L'Hôpital's Rule Goes Wrong

4.13L'Hôpital's Rule: the $0/0$ Form

4.14The $\infty/\infty$ Form and the Other Indeterminates