The real numbers, limits, the derivative and the integral, made exact.
6 chapters written · 32 definitions · 48 theorems · builds on Unit 0
The story
School calculus works with limits it never defines. This unit defines them. The starting point is a property of the real numbers that school never states: every bounded set has a least upper bound (1.1). Every theorem in the unit is that one fact, used carefully.
From it come limits of sequences with ε and N (1.2), limits of functions and continuity with ε and δ (1.3), the derivative as the best straight-line approximation (1.4), the integral as a squeeze between upper and lower sums (1.5), and series, where infinitely many terms add up to a number (1.6).
The same pattern repeats in every chapter: say what "close" means with a number, then prove that closeness can be forced. Unit 4 repeats it with distance in any space, and MATH 521 starts where this unit ends.
Every theorem in the unit, in the order it is proved. Each one is used by the ones after it. The label says whether it is on the 375/376 exams (T), an Apostol extra (A), or beyond both (B).
Theorem · BDarboux sums are integrals of step functions The sup of the lower sums equals the sup of the integrals of step functions below f, and likewise for upper sums, so Apostol's integral and the Darboux integral agree.
A mixed set of 12 quick checks drawn at random from the 56 in this unit's chapters. Each one says which chapter it comes from, so a miss tells you what to reread. Press "New set" for another mix.
Your answerYou hold the rational upper bound M=1.5 of A. The number 2M+M1 is a smaller rational upper bound (this is the Babylonian step you will meet in 1.2). Compute it, as a fraction.
21.5=43 and 1.51=32.
21.5+1.51=43+32=1217≈1.41667. It is rational, it is below 1.5, and its square is 144289>2 (since 289>288), so it is still an upper bound. Repeating the step gives smaller and smaller rational upper bounds that never stop at a least one.
Quick checkWhich of these sets has a supremum but no largest element?
[0,1) has supremum 1, but 1 is not in the set and no element is the biggest (any x<1 has 2x+1 above it). The other two sets contain their supremum 1, so 1 is their largest element.
Quick checkThe rationals are dense in the reals, yet Q is not complete. Which statement explains how both can be true?
Density is about having a rational inside every interval. Completeness is about every bounded set having its top edge in the system. A={q∈Q:q2<2} has rationals as close to 2 as we like, yet the top edge 2 is not rational.
Quick checkS={x∈Q:x2<2}. In R, what is supS, and is it in S?
In R completeness gives a supremum, 2. It is not rational, so not in S, and S has no largest element (any rational below 2 has a bigger one, by density).
Quick checkA proof of an→0 says: 'Let N=100. Then for n>N, ∣an∣<0.01.' What is wrong?
The definition starts with 'for every ε'. Showing one tolerance works proves nothing about smaller ones. Start with 'Let ε>0' and choose N in terms of ε.
Quick checkLet an=n1 and bn=n1. Both tend to 0. Why can't the quotient law tell us the limit of an/bn?
The law an/bn→a/b needs b=0. With b=0 the law says nothing, and the answer depends on the sequences: here the quotient is 1 always, but 1/n1/n2=n1→0.
Quick checkThe sequence an=(−1)n(1+n1) is bounded. Which statement about it is true?
Bounded does not give convergence ((−1)n shows that). Bolzano–Weierstrass gives a convergent subsequence: here the even terms 1+n1→1 and the odd terms −(1+n1)→−1. Two subsequences with different limits show the whole sequence diverges.
Quick checkOver the rationals, 1,1.4,1.41,1.414,… (decimal truncations of 2) is a Cauchy sequence of rationals. Does it converge in Q?
Terms after the k-th differ by less than 10−k, so it is Cauchy. In R it converges to 2. In Q there is nothing to converge to. 'Every Cauchy sequence converges' is another way of saying 'no holes'.
Your answerFor an=n1 and ε=0.1, find the smallest N such that ∣am−an∣<ε for ALL m,n>N. (Use that for m,n>N both terms lie in (0,N+11], so the worst gap is just under N+11.)
Solve N+11≤0.1.
If m,n>N then both am and an lie in (0,N+11], so ∣am−an∣<N+11. We need N+11≤0.1, so N+1≥10, so N≥9. With N=9 the terms after the 9th are 101,111,…, all in (0,0.1], and any two differ by less than 0.1. With N=8 it fails: a9=91≈0.111 and a very late term near 0 differ by more than 0.1.
Quick checkLet f(x)=x−2x2−4 for x=2 and f(2)=7. What is true at a=2?
The limit ignores the value at 2 and equals 4. Continuity needs the limit to equal f(2)=7, and 4=7. Changing the single value f(2) never changes the limit.
Quick checkf(x)=x1 is continuous on (0,1]. Does it attain a maximum there?
As x→0+, x1→∞. At x=1 the function is 1, which is its smallest value, not its largest. The extreme value theorem requires [a,b], closed and bounded; (0,1] is not closed, and the theorem's proof breaks because a subsequence could converge to 0, outside the domain.
Quick checkDefine f(x)=−1 for x<0 and f(x)=1 for x≥0. It takes the value −1 and the value 1, but never 0. Why does this not contradict the intermediate value theorem on [−1,1]?
The hypothesis is continuity on the WHOLE interval, and the jump at 0 breaks it. The signs at the ends are opposite, and the theorem is not limited to polynomials.
Quick checkWhich of these functions is uniformly continuous on its whole domain?
x2 on [0,5] is continuous on a closed bounded interval, so Heine–Cantor applies. x2 on R and x1 on (0,1) get steeper without limit, and the examples above show no single δ works. The same function x2 is uniformly continuous on a bounded interval and not on R: uniform continuity depends on the set, not only on the formula.
Quick checkWhich error E(h), in f(a+h)=f(a)+mh+E(h), is small enough for m to be the derivative?
We need hE(h)→0. For h2 the ratio is h→0. For ∣h∣ the ratio is ±1, not tending to 0. For 0.01h the ratio is the constant 0.01, which is a small number that does not tend to 0. In that case f(a+h)=f(a)+(m+0.01)h exactly, so the derivative would be m+0.01 and m would be the wrong slope. 'Small compared with h' means the ratio shrinks to 0 as h does.
The theorem proves the first. For the failure of the converse, ∣x∣ is continuous at 0 and not differentiable there. Remember the picture: continuity says the graph has no break, differentiability says it also has no corner.
Quick checkA student proves the chain rule by writing hf(g(a+h))−f(g(a))=g(a+h)−g(a)f(g(a+h))−f(g(a))⋅hg(a+h)−g(a). What is wrong?
If g is constant near a, or oscillates so that g(a+h)=g(a) for a sequence of h→0, we divide by zero. The function φ repairs exactly this: it multiplies by k instead of dividing by it, and it is defined and continuous even at k=0.
Quick checkf(x)=x3 has f′(0)=0. Does f have a maximum or minimum at 0?
Fermat's theorem says: interior extremum implies f′=0. The converse is false. x3 is negative for x<0 and positive for x>0, so 0 is neither a maximum nor a minimum.
Quick checkWhy does the mean value theorem need f continuous on the CLOSED interval [a,b]?
The proof subtracts the chord and uses that the result has a maximum and minimum. Without continuity at the endpoints that can fail: f(x)=x on [0,1) with f(1)=0 has no point where f′ equals the chord slope 0.
Quick checkFor f(x)=x on [0,1] and P={0,21,1}, what is U(f,P)?
Piece [0,21] has M1=21, piece [21,1] has M2=1, and both have width 21: U=21⋅21+1⋅21=43. The value 41 is the lower sum (use the left ends 0 and 21).
Your answerNow take f(x)=x2 on [0,1] and the same P={0,21,1}. What is the gap U(f,P)−L(f,P)?
Increasing function: the top of each piece is its right end, the bottom is its left end. Both pieces have width 21.
On [0,21] the lowest value of x2 is 0 and the highest is 41. On [21,1] they are 41 and 1. So U=21(41+1)=85 and L=21(0+41)=81, and U−L=21. (This is n1 with n=2, the pattern proved below.)
Quick checkQ refines P. Which statement does the refinement lemma give?
More cuts let each rectangle hug the curve more tightly, so outer rectangles come down and inner rectangles go up. The window between L and U can only shrink. The second option is the tempting reversal.
Your answerUse the formula Un=6n2(n+1)(2n+1) for x2 on [0,1]. What is the upper sum U100 with 100 equal pieces? (Give a decimal or a fraction.)
Put n=100: the numerator is (100+1)(200+1).
U100=6⋅10000101⋅201=6000020301=0.33835. It exceeds 31=0.33333… by 2001+600001, while the lower sum L100=0.32835 is about 2001 below it. The gap is U100−L100=1001.
Quick checkWhich feature makes the Dirichlet function fail to be integrable?
The function IS bounded and takes only two values; the example before it, the jump function, also takes two values and is integrable. The difference is that the jump has one bad point, which a thin piece can trap, while here every piece is bad.
Quick checkWhich statement about sup is used to show U(f+g,P)≤U(f,P)+U(g,P)?
If f peaks at one point and g at another, f + g never reaches the sum of the two peaks, so the inequality can be strict. Only the 'at most' is true in general.
Your answerLet F(x)=∫0x2etdt. Use part 1 and the chain rule to find F′(1).
Outer function evaluated at x2, times the derivative of x2.
F′(x)=ex2⋅2x (part 1 for the area function, then the chain rule for the upper limit x2). At x=1 this is 2e≈5.4366. Check by another route: F(x)=ex2−1, and its derivative is 2xex2.
Quick checkIf f is continuous and F(x)=∫axf(t)dt, what is F′(x)?
That is part 1: the rate at which area accumulates at x equals the height of the curve at x. The lower limit a only shifts F by a constant, which differentiation removes.
Quick checkPart 2 is used to compute ∫−11x−2dx with F(x)=−x1, giving F(1)−F(−1)=−2. What goes wrong?
Part 2 needs F continuous on [a, b] and f integrable. Here f is unbounded, so no upper sums exist, and F fails to be continuous at 0. The integrand is positive, so any 'integral' could not be negative anyway.
Your answerFind k=1∑∞2(53)k. Careful: the sum starts at k=1.
Geometric series: first term divided by 1−r.
The first term is 2⋅53=56 and the ratio is 53. The sum is 1−rfirst term=2/56/5=3. (Starting at k=0 would give 2/52=5, which includes the extra term 2.)
If the terms do NOT tend to 0, the series diverges. If they do, nothing follows. The harmonic series is the standard example of terms tending to 0 with a divergent sum.
The absolute values k21 add up (comparison test above). The middle one converges, but only because of cancellation: its absolute values give the harmonic series. The last does not converge at all (terms do not tend to 0).
Quick checkFor ∑k21 the ratio akak+1=(k+1)2k2 tends to 1. What does the ratio test tell you?
The test gives convergence only for a limit strictly below 1, and divergence only for a limit strictly above 1. At exactly 1 both ∑k1 (diverges) and ∑k21 (converges) occur, so another test is needed. The third option confuses 'each ratio is below 1' with 'the limit is below 1'.
Quick checkWhich series can be added in any order without changing its sum?
Only absolute convergence protects the sum from reordering, and ∑k21 converges. The alternating harmonic series converges only through cancellation, and Riemann's theorem says its terms can be reordered to give any sum you like. Converging is not enough.
Your answerOn [0,0.9] the functions xn converge to 0. What is sup[0,0.9]∣xn∣ for n=10? (Give a decimal.)
An increasing function takes its largest value at the right end of the interval.
xn is increasing on [0,0.9], so its largest value is at x=0.9: 0.910=0.3487. For general n the supremum is 0.9n→0, so the convergence is uniform on [0,0.9] (but not on [0,1), where the supremum is 1 for every n).
Quick checkWhich phrase describes uniform (not just pointwise) convergence?
The second is pointwise. The third is a CONSEQUENCE of uniform convergence when each fn is continuous (next theorem), not its definition: xn on [0,1) converges to the continuous function 0 pointwise but not uniformly.
Quick checkWhy does the M-test give UNIFORM convergence and not just pointwise?
Uniform means one N for all x. The tail bound is a single number, independent of x, so one N makes it below ε everywhere. Continuity plays no role in this theorem.
Quick checkA power series has radius of convergence R=2. What can you conclude at x=2?
The theorem gives absolute convergence for ∣x∣<2 and divergence for ∣x∣>2. At ∣x∣=2 the ratio test is silent and every behaviour occurs: ∑k2kxk diverges at 2, and ∑k22kxk converges there.