Learning goals and preparation

We assume Chapter 1's derivative definition and rules, continuity, and basic trigonometric functions in radians. This chapter explains how derivative information controls an entire graph or optimization problem. It also supplies carefully qualified techniques for indeterminate limits. A derivative sign or a stationary-point calculation is evidence for a specific conclusion, not a substitute for checking hypotheses.

2.1 Mean value theorems connect local and interval information

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

Consequently $f'>0$ on an interval implies strict increase there, and $f'<0$ implies strict decrease. If $|f'|\le M$ between two points, the theorem also gives the useful bound $|f(b)-f(a)|\le M|b-a|$.

Worked example 1: certify uniqueness. The equation $x+e^x=3$ has a root in $(0,2)$ because the continuous left-minus-right function is negative at zero and positive at two. Its derivative $1+e^x>0$ gives strict increase on $\mathbb R$, so there can be only one root. Existence and uniqueness require different arguments.

The interval hypotheses cannot be ignored. The function $1/x$ has negative derivative on its domain, but comparing $-1$ with $1$ crosses a missing input: $f(-1)=-1<f(1)=1$. It is decreasing on each connected domain interval, not across the gap.

2.2 Stationary points, concavity, and a complete graph analysis

A stationary point has $f'=0$. A critical input may also be an interior domain point where the derivative fails to exist. A local maximum or minimum need not be stationary if differentiability fails, and an endpoint is not an interior critical input.

For the first derivative test, let $c$ be an interior domain point, let $f$ be continuous at $c$, and let $f$ be differentiable on both sides in some punctured neighborhood of $c$. If $f\prime>0$ on the left and $f\prime<0$ on the right, then $c$ is a strict local maximum; reversing both signs gives a strict local minimum. Differentiability at $c$ itself is not required. The second derivative describes change of slope. When $f''>0$ on an interval the slope increases and the graph is convex (concave upward); when $f''<0$ it is concave downward. An inflection point requires a continuous graph point at which concavity changes. A zero or undefined second derivative alone does not prove inflection.

Proof and the continuity condition. For $x<c$ sufficiently close to $c$, the mean value theorem on $[x,c]$ gives $f(c)-f(x)=f\prime(\xi)(c-x)>0$. For $x>c$, applying it on $[c,x]$ gives $f(x)-f(c)=f\prime(\eta)(x-c)<0$. Thus every nearby other value is smaller; apply this argument to $-f$ for the minimum case. Continuity at $c$ is essential: the function with $f(x)=-|x|$ for $x\ne0$ but $f(0)=-1$ satisfies $f\prime(x)=1$ for $x<0$ and $f\prime(x)=-1$ for $x>0$, yet $0$ is a strict local minimum. The jump prevents the required mean value theorem on an interval ending at $0$.

Worked example 2: a rational graph, step by step. Let

$$f(x)=\frac{x^2}{x-1}=x+1+\frac1{x-1},\qquad x\ne1.$$

1. Domain and intercepts. The domain excludes $1$. The only intercept is $(0,0)$.
2. Asymptotes. The difference from the line $y=x+1$ is $1/(x-1)$, tending to zero as $x\to\pm\infty$, so this is an oblique asymptote. At $x\to1^-$, $f\to-\infty$; at $x\to1^+$, $f\to+\infty$.
3. Monotonicity. $f'=1-1/(x-1)^2=x(x-2)/(x-1)^2$. It is positive on $(-\infty,0)$ and $(2,\infty)$, negative on $(0,1)$ and $(1,2)$. Thus $(0,0)$ is a local maximum and $(2,4)$ a local minimum.
4. Concavity. $f''=2/(x-1)^3$ is negative left of $1$ and positive right of $1$. There is no inflection point at $1$ because it is not in the domain.

This information describes the curve without relying on a graphing window that might conceal the vertical behaviour. The local maximum value $0$ is smaller than the local minimum value $4$, which is possible because the graph has disconnected branches and these are local, not global labels.

2.3 Constrained optimization and global checks

A continuous function on a closed bounded interval attains extrema. For a differentiable objective with finitely many candidate points, evaluate all interior critical inputs and both endpoints. On an open or unbounded domain, a best value may be approached without being attained, so an additional argument is needed.

Worked example 3: minimum material for a closed cylinder. Suppose a closed cylindrical container must have fixed volume $V>0$, with radius $r>0$ and height $h>0$. Ignore thickness and joins. Volume gives $h=V/(\pi r^2)$, so surface area is

$$S(r)=2\pi r^2+2\pi rh=2\pi r^2+\frac{2V}{r}.$$

The derivative is $S'=4\pi r-2V/r^2$. Its unique zero satisfies $r^3=V/(2\pi)$. It is negative before this radius and positive afterward, so the stationary point is the global minimum over $r>0$. Alternatively $S\to\infty$ at both ends of the positive domain, together with the sign change, supplies the global check. At the minimizing radius,

$$h=V/(\pi r^2)=2r.$$

The conclusion depends on the assumptions of a closed cylinder and equal material cost per area. An open-top vessel or different side/end costs creates a different objective.

A robust setup states the objective, constraints, feasible domain, and units before differentiating. Solving an unrestricted derivative equation first and inventing constraints afterward is a frequent source of inadmissible answers.

2.4 Trigonometric limits and indeterminate forms

Sector area and radians

Use these elementary geometric facts: the unit circle has circumference $2\pi$ and enclosed area $\pi$, with the same $\pi$ in both formulas; planar area is invariant under rotations, finitely additive for regions with disjoint interiors, and monotone under inclusion.

Let $0<h<\pi/2$ be measured in radians and put $t=h/(2\pi)$. On the unit circle, $h$ is the arc length, so $t$ is the fraction of the full circumference subtended by the angle. Divide the circle into $n$ equal-angle sectors. Rotations carry each sector onto the next, so all have equal area. Their interiors are disjoint and together they fill the disk; finite additivity gives area $\pi/n$ for each. Thus a sector with rational fraction $t=m/n$ of a turn has area $m\pi/n=\pi t=h/2$.

For arbitrary $t$ in this range, choose rational numbers $0\le r_k\le t\le s_k\le1$ with $r_k,s_k\to t$. Taking all sectors from the same initial ray, the corresponding sectors are nested by inclusion, so area monotonicity and the rational case give

$$\pi r_k\le A(h)\le\pi s_k.$$

Taking $k\to\infty$ yields $A(h)=\pi t=h/2$. This establishes the sector-area formula used in the geometric comparison.

9.5 The fundamental trigonometric limit

All limits in this section use radians. For $0<h<\pi/2$, compare the areas of the inner triangle, unit-circle sector, and outer tangent triangle:

$$\tfrac12\sin h<\tfrac12h<\tfrac12\tan h.$$

The sector area is $h/2$ specifically because $h$ is in radians. Divide through and use positive cosine to obtain

$$\cos h<\frac{\sin h}{h}<1.$$

For $0<h<\pi/2$, the area comparison above also gives $0<\sin h<h$ and $\cos h>0$. Using $\sin^2 h+\cos^2 h=1$,

$$0\le 1-\cos h=\frac{\sin^2 h}{1+\cos h}\le h^2.$$

Thus $\cos h\to1$ as $h\to0^+$. Since cosine is even, the same limit holds from the left.

As $h\to0^+$, cosine tends to one, so the squeeze principle gives $\sin h/h\to1$. The quotient is an even function of $h$, giving the same left limit. The squeeze principle itself follows from the finite-limit definition: if lower and upper functions are both within any prescribed tolerance of $L$, every number between them is too.

Two consequences are

$$\lim_{h\to0}\frac{\tan h}{h}=1,\qquad \lim_{h\to0}\frac{1-\cos h}{h^2}=\frac12.$$

For the first, divide $\sin h/h$ by $\cos h$. For the second, use $1-\cos h=2\sin^2(h/2)$ to get one half times $[\sin(h/2)/(h/2)]^2$. In particular $(\cos h-1)/h\to0$.

Theorem: Cauchy's mean value theorem

If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ with $a<b$, some $c\in(a,b)$ satisfies

$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$

Proof. The auxiliary function $H(t)=[f(b)-f(a)]g(t)-[g(b)-g(a)]f(t)$ has equal endpoint values. Rolle's theorem gives $H'(c)=0$, precisely the identity. It is valid even when a denominator in a ratio formulation would be zero. Dividing to write derivative ratios requires separately checking those denominators; this is essential in proofs of L'Hopital's rule.

Learning goals and a precise one-sided setting

Identify genuine indeterminate forms, prove the rule rather than treating derivative division as algebra, and distinguish sufficient hypotheses from necessary ones. Let $f,g$ be differentiable on an open interval approaching an endpoint $b$, which may be finite or $+\infty$, from the left. Assume $g'$ is nonzero throughout a sufficiently late approach interval. The analogous right-hand approach and negative-infinity cases follow by reversing coordinates.

Cauchy's mean value theorem says that for distinct interior $x,y$, some $c$ between them satisfies

$$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(c)}{g'(c)}.$$

The denominator $g(x)-g(y)$ cannot vanish, for otherwise Rolle's theorem on $g$ would give a zero derivative between them. This also shows $g$ takes any value, including zero, at most once there. By shortening the approach interval we may therefore ensure $g\ne0$, so the desired quotient is defined.

Theorem: finite derivative-ratio limit

If $f'/g'\to L\in\mathbb R$ as the endpoint is approached, then $f/g\to L$ provided either (i) both $f,g\to0$, or (ii) $|g|\to\infty$. In case (ii), no separate assumption that $f$ is unbounded is necessary.

Proof of (i). Let $h=f-Lg$. Given $\varepsilon>0$, throughout a sufficiently late interval $|h'/g'|<\varepsilon$. Cauchy's theorem gives $|h(x)-h(y)|\le\varepsilon|g(x)-g(y)|$ for any two points there. Fix $x$ and let $y$ approach the endpoint. Since $h(y),g(y)\to0$, this yields $|h(x)|\le\varepsilon|g(x)|$, so $|f(x)/g(x)-L|\le\varepsilon$. This argument works even when the endpoint is infinity, without pretending infinity is an endpoint value of a continuous function.

Proof of (ii). Given $\eta>0$, set $\varepsilon=\eta/4$. Again take $h=f-Lg$ and a late interval where the derivative ratio has magnitude below $\varepsilon$. Fix one point $x_0$ in it. Cauchy's theorem between $x_0$ and a later $x$ gives $|h(x)-h(x_0)|\le\varepsilon|g(x)-g(x_0)|$. Consequently

$$\left|\frac{f(x)}{g(x)}-L\right| \le\frac{|h(x_0)|}{|g(x)|}+\varepsilon\left|1-\frac{g(x_0)}{g(x)}\right|.$$

For sufficiently late $x$, the first term is less than $\eta/2$ and the absolute factor in the second is less than $2$. The total is therefore less than $\eta/2+2(\eta/4)=\eta$. Since $\eta>0$ was arbitrary, the quotient tends to $L$.

Theorem: infinite derivative-ratio limits

Under the same denominator assumptions, if $f'/g'\to+\infty$, then $f/g\to+\infty$ in either case (i) or (ii); replace signs for a limit of $-\infty$.

Proof. For any real threshold $M$, on a sufficiently late interval $f'/g'>M$. Cauchy's theorem makes every incremental ratio greater than $M$. In case (i), fix $x$ and pass the other point to the endpoint to obtain $f(x)/g(x)\ge M$. Since $M$ can be chosen above any requested threshold, this proves divergence to $+\infty$. In case (ii), fix $x_0$ and write the incremental ratio as $R(x)>M$:

$$\frac{f(x)}{g(x)}=R(x)\left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}.$$

For $M>0$ and sufficiently late $x$, the multiplying factor exceeds $1/2$ and $f(x_0)/g(x)>-1$. Therefore $f(x)/g(x)>M/2-1$. Taking $M$ arbitrarily large proves divergence to $+\infty$. Applying the result to $-f$ covers the negative-infinity limit.

Merely being unbounded is not the same as tending to infinity and does not meet the stated infinite-limit hypothesis. The rule also has versions at infinity. It evaluates a limit of a quotient; it does not assert the functions $f/g$ and $f'/g'$ are equal.

Worked example 4: verify the form first. For $x\to0$, $(e^x-1)/x$ has form $0/0$, and the derivative ratio is $e^x/1\to1$, so the limit is $1$. But for $(1+x)/x$ at $x\to0^+$, the numerator tends to one, not zero. Applying the rule would falsely produce $1$; the actual quotient tends to $+\infty$.

Other indeterminate forms must be rewritten. For $x\to0^+$,

$$x\ln x=\frac{\ln x}{1/x}.$$

The numerator and denominator both tend to infinite magnitude; their derivative ratio is $-x\to0$. Hence $x\ln x\to0$. Taking logarithms then gives $x^x=\exp(x\ln x)\to1$. The logarithm step requires $x>0$, and the final exponentiation uses continuity of the exponential.

2.5 Differential error estimates and sensitivity

For small $h$, $f(a+h)-f(a)\approx f'(a)h$. The mean value theorem can turn this into a rigorous bound when the derivative is controlled over the entire segment. A first-order approximation is not itself an error bound.

Worked example 5: tolerance in a spherical volume. With $V(r)=4\pi r^3/3$, the relative differential satisfies $dV/V=3\,dr/r$. At nominal radius $r=10$ cm and tolerance $|h|\le0.01$ cm, the differential predicts a relative volume change of about $0.003$. To bound the actual change, expansion gives

$$\frac{V(10+h)-V(10)}{V(10)}=3\frac h{10}+3\left(\frac h{10}\right)^2+\left(\frac h{10}\right)^3.$$

The omitted relative terms have magnitude at most $3\cdot10^{-6}+10^{-9}$. The calculation distinguishes the predicted first-order effect from a quantified remainder.

2.6 Graded exercises

Foundation

1. Use the mean value theorem to show $|\sin b-\sin a|\le|b-a|$.
2. Classify all stationary points of $f(x)=x^3-6x^2+9x$.
3. Explain why $f(x)=x^4$ has a minimum but no inflection at zero despite $f''(0)=0$.

Core

4. Find the monotonicity intervals and local extrema of $f(x)=x+4/x$, $x\ne0$.
5. A rectangle has perimeter $20$. Find its maximum area among positive side lengths and prove global optimality.
6. Evaluate $\lim_{x\to0}(1-\cos(3x))/x^2$.
7. Evaluate $\lim_{x\to\infty}(\ln x)/x$ with the relevant limit-rule hypotheses.
8. Find $\lim_{x\to0^+}(1+x)^{1/x}$ by taking logarithms.

Synthesis

9. A closed cylinder has volume $16\pi$. Find the radius, height and surface area minimizing the equal-cost material objective.
10. Let $f(x)=x^3$. At $a>0$, derive its exact change under $h$ and the remainder after the linear term. If $|h|\le\eta a$ with $0<\eta<1$, bound the relative remainder by $3\eta^2+\eta^3$.

2.7 Full solutions

1. Sine is continuous on the segment and differentiable inside. For $a\ne b$, the theorem gives $\sin b-\sin a=\cos c(b-a)$ for some intermediate $c$. Since $|\cos c|\le1$, the bound follows. For $a=b$ it is immediate.
2. $f'=3(x-1)(x-3)$. Its signs are positive, negative, positive on the three successive intervals. Thus $x=1$ is a local maximum with value $4$, and $x=3$ a local minimum with value $0$.
3. $x^4\ge0$ with equality only at zero, proving a strict minimum. For $x\ne0$, $f''=12x^2>0$ on both sides, so concavity does not change and there is no inflection.
4. $f'=1-4/x^2=(x-2)(x+2)/x^2$. It is positive on $(-\infty,-2)$ and $(2,\infty)$, negative on $(-2,0)$ and $(0,2)$. The point $(-2,-4)$ is a local maximum; $(2,4)$ is a local minimum. Zero is outside the domain.
5. If sides are $x$ and $10-x$, feasibility is $0<x<10$. Area is $x(10-x)=25-(x-5)^2$, proving the unique global maximum $25$ at sides $5,5$.
6. Use $1-\cos3x=2\sin^2(3x/2)$. The quotient becomes $(9/2)[\sin(3x/2)/(3x/2)]^2\to9/2$.
7. Numerator and denominator both tend to infinity, are differentiable for $x>0$, and the derivative of the denominator is one. Their derivative ratio is $1/x\to0$, so the limit is zero.
8. Let $y=(1+x)^{1/x}$, $x>0$. Then $\ln y=\ln(1+x)/x$ has form $0/0$ and derivative ratio $1/(1+x)\to1$. Therefore $\ln y\to1$ and $y\to e$.
9. The minimizing radius satisfies $r^3=V/(2\pi)=8$, so $r=2$ and $h=2r=4$. Surface area is $2\pi(4)+2\pi(2)(4)=24\pi$. The derivative sign argument in Worked example 3 proves the global minimum.
10. Exact change is $(a+h)^3-a^3=3a^2h+3ah^2+h^3$. The remainder is $3ah^2+h^3$. Dividing its absolute value by $a^3$ gives at most $3(|h|/a)^2+(|h|/a)^3\le3\eta^2+\eta^3$. This is the relative remainder measured against the nominal value, not against the possibly small finite change.

Summary and onward map

Mean value theorems justify monotonicity, uniqueness and sensitivity bounds. Local critical points must be interpreted using sign changes, feasible domains and global checks. A full graph analysis separates domain, asymptotes, monotonicity and concavity. Indeterminate limit rules require verified hypotheses; logarithms can convert variable powers into quotient limits. Taylor estimates will extend the linear remainder analysis to higher orders.