Derivative depth: discontinuous derivatives and local inverse conditions

Learning goals and setting

Read differentiability as linear approximation, handle zero inner increments in the chain rule, construct a differentiable function with discontinuous derivative, and state sufficient hypotheses for a local differentiable inverse. Unless an endpoint is explicitly mentioned, the function is defined on an open neighborhood of the point.

Differentiability at $a$ means a finite limit $f'(a)$ of the difference quotient. Equivalently,

$$f(a+h)=f(a)+f'(a)h+h\varepsilon(h),\qquad\varepsilon(h)\to0.$$

Define the error quotient for nonzero $h$ and give it value zero at zero. This form is the precise content of a tangent approximation. It implies continuity by letting the increment tend to zero, but does not by itself supply a numerical error constant or a bound on the second derivative.

A vertical or one-sided infinite slope is not a finite derivative under this definition. At a domain endpoint, a one-sided derivative may be a separately named object; it must not be silently identified with an unavailable two-sided derivative.

1. A chain-rule proof that handles zero inner changes

Suppose $g$ is differentiable at $a$, $F$ is differentiable at $b=g(a)$, and the nearby composition is defined. Write

$$F(b+k)-F(b)=F'(b)k+k\eta(k),\qquad\eta(k)\to0,$$

with $\eta(0)=0$. Let $k=g(a+h)-g(a)=g'(a)h+h\varepsilon(h)$. Continuity of $g$ ensures $k\to0$. Substitution gives

$$\frac{F(g(a+h))-F(g(a))}{h} =[F'(b)+\eta(k)][g'(a)+\varepsilon(h)]\longrightarrow F'(b)g'(a).$$

This proof never divides by $k$, which may be zero for infinitely many increments. The familiar product of two difference quotients needs an extra case when its proposed denominator is zero; the error-function proof includes that case automatically.

For a quotient $f/g$, the requirement $g(a)\ne0$ plus continuity of $g$ gives a neighborhood where $|g(x)|>|g(a)|/2$. Thus the quotient is locally defined and its derivative formula can be justified. Merely writing a squared denominator in the final answer does not prove this local domain fact.

2. Differentiability does not make the derivative continuous

Define $f(0)=0$ and $f(x)=x^2\sin(1/x)$ for $x\ne0$. The difference quotient at zero is $h\sin(1/h)$, tending to zero by squeezing. Hence $f'(0)=0$. Away from zero,

$$f'(x)=2x\sin(1/x)-\cos(1/x).$$

At $x_n=1/(2n\pi)$, the derivative is minus one; at $y_n=1/((2n+1)\pi)$, it is plus one. Both sequences approach zero, so the derivative has no limit there. The original function is differentiable everywhere, while its derivative is discontinuous at zero.

This example does not imply derivatives can have arbitrary jump discontinuities; that is a different theorem question. It establishes only the precise distinction needed here: differentiability of $f$ at a point does not assert continuity of $f'$ there. The notation $C^1$ explicitly adds continuous first derivative and is stronger than mere differentiability.

3. Piecewise matching has layers

Let $f(x)=ax+b$ for $x<0$ and $f(x)=e^x$ for $x\ge0$. Its value at zero is one. Continuity requires $b=1$. With that value, the left derivative is $a$ and the right derivative one, so differentiability requires $a=1$. For those parameters the derivative equals one on the negative side and $e^x$ on the positive side, tending to one from both sides, so $f$ is $C^1$ at zero.

But the derivative's one-sided slopes at zero are zero and one, so the second derivative does not exist there. Matching a function value, a first derivative and a second derivative are separate conditions. Checking only the original branch values cannot establish any higher differentiability.

4. Prove local inverse existence before its derivative formula

Theorem: a convenient one-dimensional inverse theorem

Let $f$ be $C^1$ on an open interval around $a$ and suppose $f'(a)\ne0$. Then on a smaller interval it is one-to-one, its image is an interval around $f(a)$, and its inverse is continuously differentiable there with

$$(f^{-1})'(f(x))=\frac1{f'(x)}.$$

Proof. By derivative continuity, shrink so that $f'$ has a fixed nonzero sign and stays bounded away from zero. The mean value theorem makes $f$ strictly monotone on that interval. Continuity and the intermediate value theorem make its image an interval, and the two-sided neighborhood of $a$ gives nearby output values on both sides of $f(a)$. The inverse is continuous: images of nearby points on opposite sides of an input bracket its inverse value. For $y=f(x)$ near $y_0=f(x_0)$,

$$\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0} =\frac{x-x_0}{f(x)-f(x_0)}\longrightarrow\frac1{f'(x_0)},$$

using inverse continuity and the nonzero derivative. The derivative formula is continuous because $f'$ and the inverse are continuous and the denominator remains nonzero. Thus existence, continuity and differentiation have all been justified in that order.

These are sufficient hypotheses, not necessary ones. The function $x^3$ has a continuous inverse despite a zero derivative at zero, but its inverse has no finite derivative there. Conversely a nonzero derivative at one point, without suitable neighborhood information, does not alone guarantee local injectivity.

5. A nonzero derivative at one point can still lack local injectivity

Define $g(0)=0$ and $g(x)=x+x^2\sin(1/x^2)$ for $x\ne0$. Its derivative at zero is one because $g(h)/h=1+h\sin(1/h^2)\to1$. For nonzero inputs,

$$g'(x)=1+2x\sin(1/x^2)-\frac2x\cos(1/x^2).$$

Along $x_n=1/\sqrt{2n\pi}$ it is $1-2/x_n<0$ for sufficiently large $n$. Along $y_n=1/\sqrt{(2n+1)\pi}$ it is $1+2/y_n>0$. Both occur arbitrarily near zero.

A continuous injective function on an interval must be strictly monotone: otherwise three ordered points exhibit a peak or trough, and a height between the middle value and both endpoint values is attained twice by the intermediate value theorem. An increasing differentiable function has nonnegative derivatives, while a decreasing one has nonpositive derivatives, directly from its difference quotients. The two signs above therefore rule out injectivity on every interval around zero. The $C^1$ inverse theorem is not contradicted: derivative continuity, the hypothesis that ensured a sign on a whole neighborhood, fails.

Graded proof and diagnostic tasks

1. Hidden denominator. Identify the problem in a chain-rule proof that divides by $g(a+h)-g(a)$ without a case split. Give an inner function where that increment vanishes for every small $h$, and explain why the chain rule still holds.
2. Discontinuous derivative. Verify $f'(0)=0$ for $x^2\sin(1/x)$ extended by zero, and use the two displayed input sequences to disprove derivative continuity.
3. Piecewise orders. Find the parameters making the linear/exponential piecewise function continuous, differentiable and $C^1$ at zero. Determine whether it then has a second derivative there.
4. Inverse domain. For $x\mapsto x^2$, specify a neighborhood of a positive anchor where the theorem gives the positive-root inverse. Explain why no two-sided neighborhood of zero gives an inverse of unrestricted squaring.
5. Sufficient versus necessary. Explain how $x^3$ and the oscillatory $g$ distinguish zero derivative, nonzero derivative and the neighborhood hypotheses in an inverse theorem.
6. Approximation claim. If only $f'(a)$ is known to exist, what error statement is justified for the tangent approximation? What further hypothesis gives a quadratic numerical error bound on a specified interval?

Full solutions

1. The proposed factor can be zero even when $h\ne0$, producing an undefined $0/0$. A constant inner function $g(x)=b$ makes this happen always, while its composition with $F$ is constant and has derivative zero, equal to $F'(b)g'(a)$. The proof using $\eta(0)=0$ and the linear-error identity handles that case without division by the inner increment.
2. The derivative quotient is $h\sin(1/h)$ with magnitude at most $|h|$, so its limit is zero. At the first input sequence the sine vanishes and cosine is one, giving derivative minus one; at the second cosine is minus one, giving plus one. Different subsequential derivative limits rule out continuity at zero.
3. Continuity requires $b=1$. Under that condition differentiability requires $a=1$, and the derivative then tends to one from both sides, giving $C^1$. Its derivative has left slope zero and right slope one at zero, so the second derivative fails. There is no choice of these two parameters that repairs the second derivative while retaining the fixed exponential right branch and differentiability.
4. Around an anchor $a>0$, choose an interval contained in $(0,\infty)$, where derivative $2x$ is positive. Its inverse branch is the positive square root on the corresponding positive image interval. Any two-sided interval around zero contains both $x$ and $-x$ for some nonzero $x$, and they have the same square. Thus unrestricted squaring is not locally one-to-one there; restricting to a one-sided domain is a different statement.
5. Cubing shows a zero derivative does not prevent a continuous inverse, but that inverse need not be differentiable. The oscillatory $g$ shows a nonzero derivative at a point need not provide local injectivity when derivatives change signs arbitrarily nearby. The $C^1$ theorem supplies the neighborhood sign and continuity needed to establish the inverse before differentiating it; neither example violates its full hypotheses.
6. Differentiability gives $f(a+h)-f(a)-f'(a)h=o(h)$, so the error divided by $h$ tends to zero. It does not specify a fixed finite-error coefficient. If a continuous second derivative exists and has magnitude at most $M$ on the segment from $a$ to $a+h$, Taylor's theorem gives absolute error at most $Mh^2/2$. The bound needs that interval and derivative control, not only the tangent slope at its center.

Summary and terms

A $C^1$ function has a continuous first derivative, a stronger condition than differentiability. A linear error form states exactly what the tangent approximates. A local inverse requires one-to-one behavior on a neighborhood, not only a formally nonzero slope in a formula. Chain and inverse rules are limiting theorems with domain conditions; making those conditions explicit prevents hidden zero divisions and unsupported branch claims.

Proofs supporting the methods used in this note

Finite real limit laws and geometric sequences: the missing proofs

Work over the ordered real numbers, with the Archimedean property, positive-integer induction and the usual absolute-value triangle inequality. A sequence converges to a finite real $L$ when every $\varepsilon>0$ admits an $N$ that works for every $n\ge N$.

Uniqueness, boundedness and subsequences. If $a_n\to L,M$ and $L\ne M$, require both errors below $|L-M|/3$ at the same sufficiently large index. The triangle inequality gives $|L-M|<2|L-M|/3$, a contradiction. For boundedness, require $|a_n-L|<1$ in the tail, and take a maximum of $|L|+1$ and the finitely many initial absolute values. For a subsequence $n_1<n_2<\cdots$, induction gives $n_k\ge k$; hence every tail threshold for the original sequence also works eventually for the subsequence.

Sum, scalar and product. If $a_n\to a$, $b_n\to b$, choose both input errors below $\varepsilon/2$; then

$$|(a_n+b_n)-(a+b)|\le |a_n-a|+|b_n-b|<\varepsilon.$$

For a fixed scalar $c\ne0$ use input tolerance $\varepsilon/|c|$; $c=0$ is immediate. Choose $B\ge1$ bounding all $|b_n|$. Require

$$|a_n-a|<\frac{\varepsilon}{2B},\qquad |b_n-b|<\frac{\varepsilon}{2(|a|+1)}.$$

The common later threshold gives $|a_nb_n-ab|\le B|a_n-a|+|a||b_n-b|<\varepsilon$.

Reciprocal and quotient. If $b\ne0$, first require $|b_n-b|<|b|/2$, ensuring $|b_n|>|b|/2$ and nonzero denominators. Then

$$\left|\frac1{b_n}-\frac1b\right|\le\frac{2|b_n-b|}{|b|^2}.$$

Requiring also $|b_n-b|<\varepsilon|b|^2/2$ proves reciprocal convergence. The product law now proves $a_n/b_n\to a/b$. Any undefined initial quotients may be omitted; convergence is a tail assertion.

Order and squeeze. If $a_n\le b_n$ eventually and $a_n\to a$, $b_n\to b$, then $a\le b$: otherwise errors below $(a-b)/3$ would force $a_n>b_n$. If $a_n\le z_n\le b_n$ eventually and both outer limits are $L$, require both errors below $\varepsilon$. Then $L-\varepsilon<z_n<L+\varepsilon$, proving $z_n\to L$. This proof does not require a previously known limit of $z_n$.

For the corresponding function-limit laws, fix a common allowed approach domain $D\subseteq\mathbb R$ and an accumulation point $a$ of $D$: every punctured neighborhood of $a$ contains a point of $D$. Let $f,g:D\to\mathbb R$ have the stated finite relative-domain limits at $a$. Replace each tail threshold in the displayed estimates by a punctured-domain radius, and take the minimum of finitely many positive radii. For the product proof only local boundedness is needed: $g(x)\to b$ gives $|g(x)|\le|b|+1$ on a sufficiently small punctured $D$-neighborhood, even if $g$ is unbounded elsewhere on $D$. The same estimates prove sum, scalar, product, reciprocal and quotient laws with their stated nonzero limiting-denominator condition; the same order and squeeze estimates apply on the common approach domain. Accumulation supplies actual points when proving uniqueness or excluding an incompatible order of limiting values.

For composition let $B\subseteq\mathbb R$, let $b\in B$, and let $F:B\to\mathbb R$ be continuous at $b$ relative to $B$. Suppose $g:D\to\mathbb R$ tends to $b$ and $g(x)\in B$ throughout some sufficiently small punctured $D$-neighborhood of $a$. Given $\varepsilon>0$, continuity supplies $\eta>0$ such that $u\in B$ and $|u-b|<\eta$ imply $|F(u)-F(b)|<\varepsilon$. The limit of $g$ and the composition-domain condition supply a common positive radius forcing both conditions for every allowed approaching $x$. Thus $F(g(x))\to F(b)$. This includes inputs with $g(x)=b$ and does not divide by an inner increment.

Geometric classification. Bernoulli's bound $(1+t)^n\ge1+nt$ for $t\ge0$ follows by induction: multiply the case $n$ by the positive $1+t$ and retain the nonnegative remainder $nt^2$. If $0<q<1$, set $t=q^{-1}-1>0$. Thus

$$0<q^n\le\frac1{1+nt}\longrightarrow0.$$

For any requested tolerance, the Archimedean property supplies $N$ making the right side smaller. Hence $r^n\to0$ for $0<|r|<1$ by $|r^n|=|r|^n$; $r=0$ gives zero for every positive index. At $r=1$ the sequence is constantly one. At $r=-1$, the even and odd subsequences are one and minus one, so uniqueness and subsequence inheritance exclude convergence. If $r>1$, Bernoulli with $t=r-1$ gives $r^n\ge1+nt\to+\infty$. If $r<-1$, $|r^n|\to+\infty$ by the same argument, contradicting boundedness of any convergent real sequence. These cases exhaust all real ratios.

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.

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

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.

Definition

Function limit and continuity

Suppose $f$ is defined at points arbitrarily close to $a$ other than $a$. We write $\lim_{x\to a}f(x)=L$ if for every $\varepsilon>0$ there is $\delta>0$ such that $0<|x-a|<\delta$ and $x$ in the domain imply $|f(x)-L|<\varepsilon$. The function is continuous at $a$ when it is defined there and this limit equals $f(a)$. For endpoints use the corresponding one-sided limit.

When the domain contains points arbitrarily close to $a$ on both sides, the two-sided limit exists exactly when the left- and right-hand limits both exist and agree. Polynomial functions are continuous everywhere, rational functions where the denominator is nonzero, and sums, products, quotients with nonzero denominator, and compositions of continuous functions are continuous where defined. In radian measure the standard limit is $\lim_{x\to0}\sin x/x=1$; it supports the derivative of sine later. Limits at $+∞$ and $-∞$ describe the two directions separately; neither infinity is a real input.

Theorem

Intermediate and extreme values

If $f$ is continuous on $[a,b]$, it attains a maximum and a minimum there. If $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in[a,b]$. Closedness, boundedness and continuity matter for the extrema assertion; continuity on an open interval alone need not give extrema.

Proof

Proof idea for the intermediate value theorem

Assume $f(a)<y<f(b)$ and let $S=\{x\in[a,b]:f(x)\le y\}$. The set is nonempty and bounded above, so $c=\sup S$ exists. Continuity rules out $f(c)<y$ because then nearby points to the right would also be in $S$; it rules out $f(c)>y$ because points of $S$ arbitrarily close to $c$ would then have values above $y$. Therefore $f(c)=y$. Endpoint equality is immediate, and reversing the inequalities handles the other order.