MatIntro week 2: sequences, supremum, continuity and limits
Solved exercises from week 2 of MatIntro at KU — convergence of sequences, the supremum property, ε-δ proofs of continuity and discontinuity, and limits.
This post was AI-generated from handwritten notes in a notebook, digitised with a document scanner.
Course: MatIntro 2026, University of Copenhagen. Textbook: Tom Lindstrøm, Kalkulus. "TLO 5.1.5" means section 5.1, exercise 5.
Convergence of sequences
Method for rational expressions as : compare degrees, then divide numerator and denominator by the highest power in the denominator. The denominator then tends to a non-zero constant and the quotient rule applies.
Degrees · Limit equal · ratio of leading coefficients denominator larger · numerator larger · diverges to
Four sequences
Sequence · Terms · Result · · converges to · · converges to · · diverges to · · diverges, but bounded
Proof that has no limit: assume it tends to and take . Beyond some both and occur, so and . The triangle inequality gives
a contradiction.
Pitfall: is not constant and does not tend to . It alternates and never settles.
Pitfall: "" means the sequence diverges. is not a real number.
TLO 4.3.1
· Limit · Result · Method a) · · · divide by b) · · · divide by : c) · · · divide by : numerator , denominator d) · · ·
Pitfall (b): the limit is , not . is where the denominator goes.
Pitfall (c): the quotient rule does not apply, since the numerator does not converge. Dividing by instead sends the denominator to — a dead end.
Pitfall (d): the second fraction needs , and , so it tends to . The minus in front turns it into .
TLO 4.3.1e
by squeezing. , so .
Pitfall: dividing by hits every term: stays in the numerator. Dropping it gives instead of .
Pitfall: , not . So is not — it is not an indeterminate form at all.
TLO 4.3.2
a) (divide by ).
b) . The first fraction has degree over degree and grows like ; the second tends to . The difference diverges to .
Pitfall: is not indeterminate. Only is.
TLO 4.3.13 — is indeterminate
Find with three different outcomes for :
· · · · Limit a) · · · · b) · · · · c) · · · ·
Any other constant: , gives . This is why the quotient rule requires the denominator's limit to be non-zero.
TLO 4.3.14 — is indeterminate
Find with prescribed behaviour of . Recipe: set and , so ; check that still tends to .
a) · · · b) · · · c) · · ·
Pitfall (b): with the recipe, must tend to slower than . gives , which does not tend to .
Own example — a difference with no limit at all
, .
- and .
- has no limit (the triangle-inequality argument above).
- , so the difference is bounded and cannot diverge to .
also works, but proving has no limit takes more: if , then gives , so ; then gives ; and becomes .
Pitfall: " lies in " does not prove there is no limit — also lies in and converges. Boundedness only rules out .
The supremum property
Supremum and infimum exist for every non-empty bounded set. Maximum and minimum exist only if the value belongs to the set. A set that is not bounded above has no supremum — is not written.
TLO 2.3.1
· Set · Bounded · inf · sup a) · · both ways · · b) · · below only · · does not exist c) · · neither · — · — d) · · neither · — · — e) · · above only · — · f) · · above only · — · g) · · neither · — · —
Pitfall (b): Lindstrøm defines , so .
Pitfall (g): applying to both sides is invalid — is not monotone. The set repeats with period in both directions, so it is unbounded.
The set where sin x < 1/2 repeats forever in both directions
TLO 2.3.2
and are bounded above with . is not bounded above: can get arbitrarily close to .
TLO 2.3.3
· Set · inf · sup a) · · · (also max) b) · · · does not exist c) · · · d) · · · (also max) e) · · ·
In d), apply to all three parts; it is strictly increasing, so and survive.
TLO 2.3.5
, non-empty and bounded.
· Statement · Verdict a) · · true b) · · false c) · · true d) · · false
Counterexample to b): , gives , so . Counterexample to d): , gives , so . Also, can be empty.
Continuity with ε-δ
Definition: is continuous at if for every there is a such that implies .
Proof template:
- Let be given.
- Choose in terms of .
- Assume .
- Show .
The scratch work runs backwards to find ; the written proof runs forwards.
TLO 5.1.5a
at . , so :
For any linear : .
Input window of width δ mapped to an output window of width 2δ
Pitfall: is too large — it only gives .
Pitfall: is not . The minus in front of a bracket hits both terms.
TLO 5.1.5c
at . With :
The factor still depends on . Require ; then . Require also :
δ is the smaller of the cap 1 and ε/6
Pitfall: without the cap, fails for large . With , gives .
Pitfall: . The cross term stays and the multiplies all three terms.
TLO 5.1.5b
at . .
is the input distance, controlled by . is a stretch factor that is only bounded because has been tied to first. The cap is arbitrary: gives and .
Left: the cap |x−3| < 1/2 and the final δ-window on the graph of x². Right: the stretch factor |x+3| stays below 6.5 on the cap
Pitfall: the cap goes on , never on the other factor. is impossible near .
TLO 5.1.5d
at . . Cap ; all terms of increase with , so the worst case is , giving :
Pitfall: . The formula is .
TLO 5.1.5e
at . . Cap gives , so (a decreasing factor is worst at the left edge):
Pitfall: with a strict sign fails at (). Use .
TLO 5.1.7b
at . is continuous, is continuous at , so is continuous at by the composition theorem 5.1.6. is continuous at . The product is continuous by 5.1.4.
Pitfall: is not a basic function; it is composed with . Both 5.1.4 and 5.1.6 are needed.
TLO 5.1.7c
at . Numerator and denominator are continuous, and , so the quotient is continuous by 5.1.4.
Pitfall: no composition here — only division, so 5.1.4, not 5.1.6. The quotient rule needs the denominator at the point, stated explicitly.
Proving discontinuity
Negation of continuity: there exists such that for every there exists with and .
· continuous · discontinuous · given — all · chosen — one · chosen — one · given — all · all with · one, constructed
TLO 5.1.6a
for , for . ; from the left ; jump .
. For arbitrary , set and . Then and .
Pitfall: must be strictly below the jump. fails: from the left, .
Pitfall: is fixed before is given. depends on and proves nothing.
Pitfall: must be on the jumping side — the branch without the equality sign.
Own exercise — jump to the right
for , for . , right limit , jump . Choose and :
so .
Pitfall: does not give . Remove the absolute value by a sign argument: , so .
TLO 5.1.9a
is a polynomial — continuous everywhere, no points of discontinuity.
TLO 5.1.9b
for , for . Only is in question. , right limit , jump . , on the right branch:
so . Discontinuous at only.
Pitfall: multiplying by flips the inequality: becomes .
TLO 5.1.9c
for , . There is no jump — oscillates between and near .
. For arbitrary , choose with (Archimedes' principle) and set . Then and
Discontinuous at only.
Pitfall: is evaluated at , not at . , which is not (for it is ).
Jump function — not continuous, and no limit at all
for , for .
Not continuous at : , gives . No is needed — is constant on .
No limit for any : take . For every , the points have values and . If both were within of , then — impossible. So redefining cannot make continuous. (Equivalently: the one-sided limits are from the left and from the right.)
Pitfall: the negation starts with "there exists ", not "for every ", and requires , not .
Limits of functions
TLO 5.4.1a
The quotient rule (5.4.3) applies because the denominator's limit is .
TLO 5.4.1c
using , , .
At θ = π the point on the unit circle is (−1, 0): cos π = −1, sin π = 0, tan π = 0. At θ = π/2, tan is undefined
Pitfall: , not . The point at angle is ; cosine is the -coordinate. Setting gives instead of .
Pitfall: is undefined where , at — not at or , where .
Pitfall: , not . Check: must equal , and .
TLO 5.4.2b
Prove from the definition: for every find such that implies .
Same estimate as TLO 5.1.5b: and
The only difference from continuity is : the point itself is excluded.
Pitfall: cannot be defined in terms of itself — is circular. The first entry is the cap .
Pitfall: the definition is an implication ("if … then"), not two statements joined by "and".
Pitfall: is false — multiplying by makes it larger.
TLO 5.4.3a
cancels because in a limit. The denominator tends to .
Pitfall: as , factor out the lowest power. Dividing by the highest power turns into .
Limit · Method · divide by the highest power · factor out the lowest power and cancel
Complex roots
TLO 3.4.2
, so is a square root of . The other is : if then , so by the zero-product rule.
TLO 3.4.5
. , , so and . First root: ; add twice:
Pitfall: does not reduce to . , so . Angles are reduced modulo .
TLO 3.4.11b
: , , . , and :
Both roots are purely imaginary.
Pitfall: , not , so . The shortcut (Theorem 3.4.8 iii) is stated for real coefficients; the general formula is .
Pitfall: — split off the largest square factor ( does not help).
Intermediate value theorem
on ; find with .
is continuous on the closed interval, , , and , so such a exists. Solve , i.e. : or . Only , so .
Pitfall: is where takes the value , not where the graph crosses the -axis — has no real roots.