Let fₙ(x) = xⁿ. Which statements are true?

(fₙ) converges pointwise on [0, 1]

(fₙ) converges uniformly on [0, 1/2]

(fₙ) converges pointwise on [0, 1]

(fₙ) converges uniformly on [0, 1]

(fₙ) converges uniformly on [0, 1/2]

the pointwise limit is continuous on [0, 1]

The sequence fₙ(x) = nx/(1 + n²x²) on [0, 1]

converges to 0 pointwise but not uniformly

converges to a discontinuous function

For the power series Σ_{n≥1} xⁿ/n, which statements are true?

it converges uniformly on [−1, 0]

Let fₙ(x) = sin(nx)/√n on ℝ. Which statement is correct?

fₙ → 0 uniformly on ℝ, but (fₙ′(0)) diverges

fₙ → 0 uniformly, so fₙ′ → 0 uniformly

fₙ does not converge uniformly on ℝ

Let f ∈ C[0, 1] satisfy ∫₀¹ f(x) xⁿ dx = 0 for every integer n ≥ 0. Then

f(x) = 0 for every x ∈ [0, 1]

f vanishes only at the rational points

Which of the following families of functions on [0, 1] are equicontinuous?

{f : f differentiable on [0, 1] with |f′(x)| ≤ 1}

Let f: ℝ² → ℝ², f(x, y) = (eˣ cos y, eˣ sin y). Which statements are true?

the Jacobian determinant of f never vanishes

f is locally invertible near every point of ℝ²

the Jacobian determinant of f never vanishes

f is locally invertible near every point of ℝ²

Sequences and Series of Functions, Power Series, and the Inverse and Implicit Function Theorems

The second Real Analysis chapter is about limits of functions. Pointwise convergence is too weak to carry continuity, integrals or derivatives to the limit; uniform convergence carries the first two and, with a condition on the derivatives, the third. The Weierstrass M-test is the everyday tool for proving a series converges uniformly; Dini’s theorem, the Ascoli–Arzelà theorem and the Weierstrass approximation theorem are the structural results the paper names. Power series are the best-behaved series of all — they converge uniformly on every compact subset of their disc — and their radius of convergence is a guaranteed numerical question. The chapter ends with the calculus of several variables at MA depth: the derivative as a linear map, and the inverse and implicit function theorems, whose conclusions are local and whose examples show that local is all they give.

1. Pointwise and uniform convergence

fₙ → f pointwise on E if fₙ(x) → f(x) for each x; uniformly if Mₙ = sup_x∈E |fₙ(x) − f(x)| → 0. The Cauchy criterion: (fₙ) converges uniformly iff for every ε > 0 there is N with |fₙ(x) − fₘ(x)| < ε for all m, n ≥ N and all x. To test uniform convergence, find the pointwise limit, then compute or bound Mₙ, usually by calculus.

Four sequences to know
fₙ(x)	Pointwise limit	sup \|fₙ − f\|	Uniform?
xⁿ on [0, 1]	0 for x < 1, 1 at x = 1	1 (take x close to 1)	no; yes on [0, a] for a < 1, where it is aⁿ
x/(1 + nx²) on ℝ	0	1/(2√n), at x = 1/√n	yes
nx/(1 + n²x²) on [0, 1]	0	1/2, at x = 1/n	no
sin(nx)/√n on ℝ	0	1/√n	yes, but fₙ′(0) = √n diverges

2. What uniform convergence preserves

Continuity. A uniform limit of continuous functions is continuous. Contrapositive: xⁿ on [0, 1] has a discontinuous limit, so the convergence cannot be uniform.
Integrals. If fₙ → f uniformly on [a, b] and each fₙ is Riemann integrable, so is f and ∫fₙ → ∫f. Pointwise is not enough: fₙ(x) = nx(1 − x²)ⁿ → 0 on [0, 1], but ∫₀¹ fₙ = n/(2(n + 1)) → 1/2.
Derivatives. If each fₙ is differentiable on [a, b], (fₙ′) converges uniformly and (fₙ(x₀)) converges for one x₀, then fₙ converges uniformly to some f with f′ = lim fₙ′. Uniform convergence of fₙ alone is not enough: sin(nx)/√n.
Dini’s theorem. If continuous fₙ on a compact K converge monotonically and pointwise to a continuous f, the convergence is uniform. All four hypotheses matter; xⁿ on [0, 1] fails only continuity of the limit.

Weierstrass M-test. If |fₙ(x)| ≤ Mₙ for all x ∈ E and ΣMₙ < ∞, then Σfₙ converges absolutely and uniformly on E. So Σ sin(nx)/n² and Σ 1/(n² + x²) converge uniformly on ℝ and define continuous functions. Σ xⁿ/n! converges on ℝ but not uniformly on ℝ (its terms at x = n are not small), only on bounded sets.

3. Power series and the radius of convergence

Σaₙ(x − c)ⁿ has a radius of convergence R ∈ [0, ∞] given by Cauchy–Hadamard, 1/R = limsup |aₙ|^1/n, or by the ratio test 1/R = lim |aₙ₊₁/aₙ| when that limit exists. The series converges absolutely for |x − c| < R, diverges for |x − c| > R, and converges uniformly on every closed interval [c − r, c + r] with r < R. Term-by-term differentiation and integration keep the same R.

Radii computed
Series	Computation	R
Σ n! xⁿ	ratio (n + 1) → ∞	0
Σ xⁿ/n!	ratio 1/(n + 1) → 0	∞
Σ (3ⁿ + 4ⁿ) xⁿ	(3ⁿ + 4ⁿ)^1/n → 4	1/4
Σ (n/(n + 1))^n² xⁿ	\|aₙ\|^1/n = (1 + 1/n)⁻ⁿ → 1/e	e
Σ 2ⁿ x²ⁿ	converges iff 2x² < 1	1/√2

⚠️ The end points must be tested separately

All three of Σxⁿ, Σxⁿ/n and Σxⁿ/n² have R = 1. The first converges at neither end point, the second at x = −1 only (alternating harmonic), the third at both. Abel’s theorem: if the series converges at x = R, the convergence is uniform on [0, R] and the sum is continuous there; so Σ(−1)ⁿ⁺¹/n = lim_x→1⁻ log(1 + x) = log 2.

4. Equicontinuity, Ascoli–Arzelà and Weierstrass approximation

A family F ⊆ C(K) is equicontinuous if for every ε > 0 there is one δ working for every f ∈ F at once. Ascoli–Arzelà: for K compact metric, F ⊆ C(K) has compact closure in the sup norm if and only if F is uniformly bounded and equicontinuous; so every bounded equicontinuous sequence has a uniformly convergent subsequence. A family with a common Lipschitz bound, e.g. {f : |f′| ≤ 1, |f(0)| ≤ 1} on [0, 1], qualifies. {sin(nx)} and {xⁿ} on [0, 1] are uniformly bounded but not equicontinuous (near 0, respectively near 1).

Weierstrass approximation theorem: every continuous f on [a, b] is a uniform limit of polynomials — for instance of its Bernstein polynomials Bₙf(x) = Σ f(k/n) C(n, k) xᵏ(1 − x)ⁿ⁻ᵏ on [0, 1]. Consequence: if f ∈ C[0, 1] and ∫₀¹ f(x)xⁿ dx = 0 for every n ≥ 0, then ∫f·p = 0 for every polynomial p, hence ∫f² = 0 and f ≡ 0. The theorem needs a compact interval: a uniform limit of polynomials on all of ℝ is itself a polynomial, so eˣ is not one.

5. Derivatives in several variables; the inverse and implicit function theorems

For f: U ⊆ ℝⁿ → ℝᵐ the derivative Df(a) is the linear map with f(a + h) = f(a) + Df(a)h + o(‖h‖); its matrix is the Jacobian (∂fᵢ/∂xⱼ), and the chain rule is D(g ∘ f)(a) = Dg(f(a))·Df(a). If the partials exist and are continuous on U, f is C¹ there.

Inverse function theorem. If f: U → ℝⁿ is C¹ and det Df(a) ≠ 0, there are open V ∋ a and W ∋ f(a) such that f: V → W is a bijection with C¹ inverse, and D(f⁻¹)(f(a)) = Df(a)⁻¹. The conclusion is local: f(x, y) = (eˣ cos y, eˣ sin y) has det Df = e^2x ≠ 0 everywhere, yet f(x, y + 2π) = f(x, y), so f is not injective on ℝ², and it misses (0, 0).
Implicit function theorem. If F: ℝⁿ × ℝᵐ → ℝᵐ is C¹, F(a, b) = 0 and the m × m matrix D_yF(a, b) is invertible, then near (a, b) the solutions of F(x, y) = 0 are exactly y = g(x) for a unique C¹ g with g(a) = b, and Dg = −(D_yF)⁻¹DₓF. For one equation in two variables, dy/dx = −Fₓ/F_y.

⚠️ A zero Jacobian does not forbid an inverse

The theorems give sufficient conditions only. f(x) = x³ has f′(0) = 0, yet it is a bijection of ℝ with continuous inverse ∛x; what fails is differentiability of the inverse at 0. Likewise x² + y² = 1 cannot be solved for y near (1, 0) because F_y = 2y = 0 there, but it can be solved for x = √(1 − y²), since Fₓ = 2 ≠ 0.

Key takeaways

Uniform convergence means sup |fₙ − f| → 0; compute that supremum, usually at a critical point such as x = 1/n or 1/√n.
Uniform limits keep continuity and integrals; derivatives need uniform convergence of fₙ′ plus convergence at one point.
The M-test proves uniform convergence; power series converge uniformly on compact subsets of the open interval of convergence, and end points need their own test.
Ascoli–Arzelà: bounded plus equicontinuous gives a uniformly convergent subsequence; Weierstrass: polynomials are dense in C[a, b].
Inverse and implicit function theorems need a C¹ map and an invertible derivative, and give only local conclusions.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Let fₙ(x) = xⁿ. Which statements are true?
1. (fₙ) converges pointwise on [0, 1]
2. (fₙ) converges uniformly on [0, 1]
3. (fₙ) converges uniformly on [0, 1/2]
4. the pointwise limit is continuous on [0, 1]
Show answer
Answer: A — (fₙ) converges pointwise on [0, 1]; C — (fₙ) converges uniformly on [0, 1/2]
The limit is 0 on [0, 1) and 1 at x = 1: pointwise convergence holds, and the limit is discontinuous, so (4) fails and the convergence on [0, 1] cannot be uniform (a uniform limit of continuous functions is continuous). On [0, 1/2], sup |xⁿ| = 2⁻ⁿ → 0, so (3) holds.
For fₙ(x) = x/(1 + nx²) on [0, ∞), the value of sup_x≥0 |f₁₆(x)| is ____.
Numerical answer — type the value.
Show answer
Answer: 0.125
fₙ′(x) = (1 − nx²)/(1 + nx²)² vanishes at x = 1/√n, where fₙ = (1/√n)/2 = 1/(2√n). For n = 16 this is 1/8 = 0.125. Since 1/(2√n) → 0, fₙ → 0 uniformly on [0, ∞).
The sequence fₙ(x) = nx/(1 + n²x²) on [0, 1]
1. converges to 0 pointwise but not uniformly
2. converges to 0 uniformly
3. does not converge at x = 0
4. converges to a discontinuous function
Show answer
Answer: A — converges to 0 pointwise but not uniformly
fₙ(0) = 0 and for x > 0, fₙ(x) ≤ nx/(n²x²) = 1/(nx) → 0, so the limit is the continuous function 0. But fₙ(1/n) = 1/2 for every n, so sup |fₙ| = 1/2 ↛ 0. A continuous limit does not make convergence uniform; Dini fails here because the sequence is not monotone.
Which series converge uniformly on the whole real line ℝ?
1. Σ sin(nx)/n²
2. Σ xⁿ/n!
3. Σ 1/(n² + x²)
4. Σ cos(nx)/n
Show answer
Answer: A — Σ sin(nx)/n²; C — Σ 1/(n² + x²)
(1) |sin(nx)/n²| ≤ 1/n² and (3) 1/(n² + x²) ≤ 1/n², so the M-test applies. (2) converges pointwise to eˣ but not uniformly: the n-th term at x = n is nⁿ/n! ≥ 1, so the terms do not tend to 0 uniformly. (4) diverges at x = 0, where it is the harmonic series.
The radius of convergence of the power series Σ_n≥1 (n/(n + 1))^n² xⁿ, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.72
By Cauchy–Hadamard, |aₙ|^1/n = (n/(n + 1))ⁿ = 1/(1 + 1/n)ⁿ → 1/e, so R = e = 2.718…, i.e. 2.72. The ratio test is awkward here; the root test is the natural one because the exponent is n².
The radius of convergence of Σ_n≥0 (3ⁿ + 4ⁿ) xⁿ is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
(3ⁿ + 4ⁿ)^1/n = 4(1 + (3/4)ⁿ)^1/n → 4, so R = 1/4. The larger base dominates; adding the two radii 1/3 and 1/4 or taking 1/7 are the usual slips.
For the power series Σ_n≥1 xⁿ/n, which statements are true?
1. its radius of convergence is 1
2. it converges at x = −1
3. it converges at x = 1
4. it converges uniformly on [−1, 0]
Show answer
Answer: A — its radius of convergence is 1; B — it converges at x = −1; D — it converges uniformly on [−1, 0]
(1) (1/n)^1/n → 1. (2) At x = −1 it is the alternating harmonic series, convergent. (3) At x = 1 it is the harmonic series, divergent. (4) For x = −t, t ∈ [0, 1], it is alternating with decreasing terms tⁿ/n, so the tail after N terms is at most t^N+1/(N + 1) ≤ 1/(N + 1), uniformly in t.
Let fₙ(x) = sin(nx)/√n on ℝ. Which statement is correct?
1. fₙ → 0 uniformly on ℝ, but (fₙ′(0)) diverges
2. fₙ → 0 uniformly, so fₙ′ → 0 uniformly
3. fₙ does not converge uniformly on ℝ
4. fₙ′(x) → 0 for every x
Show answer
Answer: A — fₙ → 0 uniformly on ℝ, but (fₙ′(0)) diverges
|fₙ| ≤ 1/√n → 0, so the convergence is uniform. But fₙ′(x) = √n cos(nx), and fₙ′(0) = √n → ∞. Uniform convergence of the functions says nothing about their derivatives; the term-by-term theorem needs uniform convergence of fₙ′.
The value of lim_n→∞ ∫₀¹ n x (1 − x²)ⁿ dx is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
With u = 1 − x², ∫₀¹ x(1 − x²)ⁿ dx = 1/(2(n + 1)), so the integral is n/(2(n + 1)) → 1/2. The integrand tends to 0 at every x ∈ [0, 1], so the limit of the integrals is not the integral of the limit; the convergence is not uniform (the peak near x = 1/√(2n + 1) grows like √n).
Let f ∈ C[0, 1] satisfy ∫₀¹ f(x) xⁿ dx = 0 for every integer n ≥ 0. Then
1. f(x) = 0 for every x ∈ [0, 1]
2. f vanishes only at the rational points
3. f is a non-zero constant
4. nothing can be concluded
Show answer
Answer: A — f(x) = 0 for every x ∈ [0, 1]
By linearity ∫f·p = 0 for every polynomial p. Weierstrass gives polynomials pₖ → f uniformly, so ∫f² = lim ∫f·pₖ = 0; f² is continuous and non-negative, so f ≡ 0. A non-zero constant c fails at n = 0, where ∫c = c.
Which of the following families of functions on [0, 1] are equicontinuous?
1. {sin(nx) : n ∈ ℕ}
2. {f : f differentiable on [0, 1] with |f′(x)| ≤ 1}
3. {xⁿ : n ∈ ℕ}
4. {sin(x + n) : n ∈ ℕ}
Show answer
Answer: B — {f : f differentiable on [0, 1] with |f′(x)| ≤ 1}; D — {sin(x + n) : n ∈ ℕ}
(2) and (4) share the Lipschitz constant 1 (mean value theorem; |cos| ≤ 1), so δ = ε works for all members. (1) fails near 0: sin(n · π/(2n)) − sin 0 = 1 while π/(2n) → 0. (3) fails near 1: 1 − (1 − 1/n)ⁿ → 1 − 1/e at distance 1/n.
Let f: ℝ² → ℝ², f(x, y) = (eˣ cos y, eˣ sin y). Which statements are true?
1. the Jacobian determinant of f never vanishes
2. f is locally invertible near every point of ℝ²
3. f is injective on ℝ²
4. f maps ℝ² onto ℝ²
Show answer
Answer: A — the Jacobian determinant of f never vanishes; B — f is locally invertible near every point of ℝ²
Df = [[eˣ cos y, −eˣ sin y], [eˣ sin y, eˣ cos y]] with determinant e^2x > 0, so (1) holds and the inverse function theorem gives (2). But f(x, y) = f(x, y + 2π), so (3) fails, and ‖f(x, y)‖ = eˣ > 0, so (0, 0) is never attained and (4) fails. This is the complex exponential e^z.
The equation x³ + y³ + xy − 3 = 0 defines y as a C¹ function of x near (1, 1). The value of dy/dx at x = 1 is ____.
Numerical answer — type the value.
Show answer
Answer: -1
F(1, 1) = 1 + 1 + 1 − 3 = 0 and F_y = 3y² + x = 4 ≠ 0, so the implicit function theorem applies with dy/dx = −Fₓ/F_y = −(3x² + y)/(3y² + x) = −4/4 = −1. Typed, the answer is -1.