M2: Analysis II

Question 1

Proposition

Limit Points via Sequences

Answer 1

p ∈ ℝ is a limit point of E ⊆ ℝ iff
∃ sequence (pₙ) pₙ ∈ E\{p}: pₙ → p

Question 2

Proof

(Limit Points via Sequences)
p ∈ ℝ is a limit point of E ⊆ ℝ iff
∃ sequence (pₙ) pₙ ∈ E\{p}: pₙ → p

2 points

Answer 2

• (⇒) Sandwich with δ = 1/n
• (⇐) Use convergence definition

Question 3

Proposition

Limits via Sequences

Answer 3

If f: E → ℝ, p is a limit point of E, L ∈ ℝ then f(x) → L as x → p iff
For every sequence (pₙ) pₙ ∈ E\{p} pₙ → p
The sequence f(pₙ) → L as n → ∞.

Question 4

Proof

(Limits via Sequences)
If f: E → ℝ, p is a limit point of E, L ∈ ℝ then f(x) → L as x → p iff
For every sequence (pₙ) pₙ ∈ E{p} pₙ → p
The sequence f(pₙ) → L as n → ∞.

3 points

Answer 4

• (⇒) Use δ in limit definition as ε in convergence definition
• (⇐) Use contrapositive
• Let δ = 1/n to find sequence

Question 5

Theorem

AOL for Functions

4 points

Answer 5

Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
Then f(x) ± g(x) → a ± b
f(x)g(x) → ab
f(x)/g(x) → a/b if b ≠ 0
|f(x)| → |a|
as x → p

Question 6

Proof

(AOL for Functions)
Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
Then f(x) ± g(x) → a ± b
f(x)g(x) → ab
f(x)/g(x) → a/b if b ≠ 0
|f(x)| → |a|
as x → p

2 points

Answer 6

• Pick pₙ ∈ E\{p} pₙ → p
• Use Analysis I AOL

Question 7

Theorem

Extended AOL

4 points

Answer 7

Let p ∈ ℝ be a limit point of E ⊆ ℝ
f: E → ℝ and g: E → ℝ are functions s.t. f(x) → a and g(x) → b as x → p
a and/or b = ±∞
Then f(x) ± g(x) → a ± b except for ∞ - ∞ or -∞ + ∞
f(x)g(x) → ab except for a = 0 or b = 0
f(x)/g(x) → a/b if b ≠ 0 except ∞/∞ or b = 0
|f(x)| → |a|
as x → p

Question 8

Theorem

Limits Preserve Weak Inequalities

Answer 8

f(x) → a, g(x) → b as x → p
∀ x: f(x) ≤ g(x) ⇒ a ≤ b

Question 8

Theorem

Sandwiching

Answer 8

f(x), g(x) → a as x → p
f(x) ≤ h(x) ≤ g(x) ⇒ h(x) → a as x → p

Question 8

Theorem

Algebra of Continuous Functions

Answer 8

If f, g: E → ℝ, f, g continouous at p ∈ E
⇒ f(x) ± g(x), f(x)g(x), |f(x)| are continuous at p
f(x)/g(x) is continuous at x = p provided g(p) ≠ 0

Question 9

Proof

(Algebra of Continuous Functions)
If f, g: E → ℝ, f, g continuous at p ∈ E
⇒ f(x) ± g(x), f(x)g(x), |f(x)| are continuous at p
f(x)/g(x) is continuous at x = p provided g(x) ≠ 0

2 points

Answer 9

• Immediate for p isolated
• Use AOL for p limit point

Question 9

Theorem

Continuous Functions Commute with Limits

Answer 9

If f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
p ∈ ℝ is a limit point of E (or ±∞ E unbounded)
f(x) → L ∈ E’ as x → p, g continuous at L
Then lim[x→p] g(f(x)) = g(lim[x → p] f(x)) = g(L)

Question 10

Proof

(Continuous Functions Commute with Limits)
If f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
p ∈ ℝ is a limit point of E (or ±∞ E unbounded)
f(x) → l ∈ E’ as x → p, g continuous at l
Then lim[x→p] g(f(x)) = g(lim[x → p] f(x)) = g(l)

2 points

Answer 10

• Use continuiuty and limit definitions
• Use δ in first definition as ε in second

Question 11

Theorem

Composition of Continuous Functions

Answer 11

f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
If f(x) is continuous at p and g(x) is continuous at f(p)
Then g(f(x)) is continuous at p

Question 12

Proof

(Composition of Continuous Functions)
f: E → ℝ, g: E’ → ℝ, f(E) ⊆ E’
If f(x) is continuous at p and g(x) is continuous at f(p)
Then g(f(x)) is continuous at p

2 points

Answer 12

• Immediate for p isolated
• Use continuous functions commute with limits for p limit point

Question 13

Theorem

Boundedness Theorem

Answer 13

Suppose f: [a, b] → ℝ is continuous (a < b)
Then f is bounded on [a, b] and:
∃ x₁ ∈ [a, b]: f(x₁) = sup f
∃ x₂ ∈ [a, b]: f(x₂) = inf f

Question 14

Proof

(Boundedness Theorem)
Suppose f: [a, b] → ℝ is continuous (a < b)
Then f is bounded on [a, b] and:
∃ x₁ ∈ [a, b]: f(x₁) = sup f
∃ x₂ ∈ [a, b]: f(x₂) = inf f

6 points

Answer 14

• Suppose f has no upper bound
• Find a sequence (pₙ) s.t. f(pₙ) ≥ n
• Use Bolzano-Weierstrass and continuity to show f(pₛ₍ₙ₎) converges
• Contradiction as f(pₛ₍ₙ₎) ≥ sₙ
• 1/(sup f - f(x)) bounded on [a, b]
• Find contradiction of sup definition

Question 15

Theorem

Intermediate Value Theorem IVT

Answer 15

Assume f: [a, b] → ℝ is continuous and f(a) ≤ c ≤ f(b) [or f(a) ≥ c ≥ f(b)]
Then ∃ ξ ∈ [a, b]: f(ξ) = c

Question 16

Proof

(Intermediate Value Theorem IVT)
Assume f: [a, b] → ℝ is continuous and f(a) ≤ c ≤ f(b) [or f(a) ≥ c ≥ f(b)]
Then ∃ ξ ∈ [a, b]: f(ξ) = c

2 points

Answer 16

• Construct [aₙ, bₙ] s.t. f(aₙ) ≤ c ≤ f(bₙ) and aₙ increasing, bₙ decreasing, bₙ - aₙ → 0 by divide and conquer
• Use continuity to show f(L) = c where aₙ, bₙ → L

Question 17

Theorem

Continuous Inverse Function Theorem CIFT

Answer 17

If I is an interval, f: I → ℝ strictly monotonic, continuous
Then f(I) is an interval and f⁻¹: f(I) → I is also strictly monotonic and continuous

Question 18

Proof

(Continuous Inverse Function Theorem CIFT)
If I is an interval, f: I → ℝ strictly monotonic, continuous
Then f(I) is an interval and f⁻¹: f(I) → I is also strictly monotonic and continuous

5 points

Answer 18

• Pick x = f(a), z = f(b) where a, b ∈ I
• Use IVT on y where x ≤ y ≤ z
• Show f(x) < f(y) ⇒ f⁻¹(f(x)) < f⁻¹(f(y)) as f strictly increasing and by trichotomy
• Set δ = min(f(x) - f(x - ε), f(x + ε) - f(x)) > 0
• Show |f⁻¹(q) - f⁻¹(p)| < ε where q ∈ f(I) and p = f(x)

Question 19

Proposition

Uniform Continuity via Sequences

Answer 19

f: E → ℝ is uniformly continuous
⇔ ∀ xₙ, yₙ ∈ E, |xₙ - yₙ| → 0: |f(xₙ) - f(yₙ)| → 0

Question 20

Proof

(Uniform Continuity via Sequences)
f: E → ℝ is uniformly continuous
⇔ ∀ xₙ, yₙ ∈ E, |xₙ - yₙ| → 0: |f(xₙ) - f(yₙ)| → 0

4 points

Answer 20

• (⇒) Use uniform continuity and convergence definitions
• (⇐) Use contradiction
• Let δ = 1/n
• Construct (xₙ), (yₙ) s.t. |xₙ - yₙ| < 1/n but |f(xₙ) - f(yₙ)| ≥ ε

Question 21

Theorem

Continuity Implies Uniform Continuity on Closed Bounded Intervals

Answer 21

If f: [a, b] → ℝ is continuous then it is uniformly continuous

Question 22

# Proof (Continuity Implies Uniform Continuity on Closed Bounded Intervals) If f: [a, b] → ℝ is continuous then it is uniformly continuous | 4 points

Answer 22

* Suppose f is not uniformly continuous and find such sequences * Use Bolzano-Weierstrass * Use continuity of f at p where xₙ₍ₖ₎ → p * Use triangle inequality to show |f(xₙ₍ₖ₎) - f(yₙ₍ₖ₎)| → 0

Question 23

# Theorem Uniform Limits Preserve Continuity

Answer 23

If fₙ, f: E → ℝ and fₙ is continuous for each n and fₙ → f uniformly on E Then f is continuous

Question 24

# Proof Uniform Limits Preserve Continuity If fₙ, f: E → ℝ and fₙ is continuous for each n and fₙ → f uniformly on E Then f is continuous | 2 points

Answer 24

* |f(x) - f(p)| ≤ |f(x) - fₙ(x)| + |fₙ(x) - fₙ(p)| + |fₙ(p) - f(p)| * Use uniform convergence and continuity to bound each term

Question 25

# Theorem Cauchy Convergence Criterion for Uniform Convergence of Sequences

Answer 25

Let fₙ: E → ℝ be a sequence of fₙs Then fₙ converges uniformly to some f ⇔ ∀ ε > 0: ∃ N: ∀ n, m ≥ N: ∀x: |fₙ(x) - fₘ(x)| < ε

Question 26

# Proof (Cauchy Convergence Criterion for Uniformly Convergent Series) Let fₙ: E → ℝ be a sequence of fₙs Then fₙ converges uniformly to some f ⇔ ∀ ε > 0: ∃ N: ∀ n, m ≥ N: ∀x: |fₙ(x) - fₘ(x)| < ε | 3 points

Answer 26

* (⇒) Use triangle inequality on |fₙ(x) - fₘ(x)| * (⇐) Fix x to show fₙ(x) → f(x) for some x * Let m → ∞ in Cauchy convergence definition

Question 27

# Theorem Weierstrass M-Test

Answer 27

Suppose uₖ: E → ℝ are such that ∀ k: ∃ Mₖ: ∀ x: |uₖ(x)| ≤ Mₖ and Σ Mₖ converges Then Σ uₖ converges uniformly on E

Question 28

# Proof (Weierstrass M-Test) Suppose uₖ: E → ℝ are such that ∀ k: ∃ Mₖ: ∀ x: |uₖ(x)| ≤ Mₖ and Σ Mₖ converges Then Σ uₖ converges uniformly on E | 1 point

Answer 28

* Use Cauchy convergence definition

Question 29

# Theorem Uniform Convergence and Continuity of Power Series

Answer 29

Suppose Σ aₖxᵏ has ROC R > 0 and suppose 0 < ρ < R Then Σ aₖxᵏ converges uniformly on {x: |x| ≤ ρ}

Question 30

# Proof (Uniform Convergence and Continuity of Power Series) Suppose Σ aₖxᵏ has ROC R > 0 and suppose 0 < ρ < R Then Σ aₖxᵏ converges uniformly on {x: |x| ≤ ρ} | 1 point

Answer 30

* Apply M-Test with Mₖ = |aₖ|ρᵏ

Question 31

# Theorem Algebraic Properties of Differentiation | 3 points

Answer 31

Suppose f, g: E → ℝ are differentiable at x₀ (a) (Linearity) a, b constants a, b ∈ ℝ af(x) + bg(x) is differentiable at x₀ with derivative af'(x₀) + bg'(x₀) (b) (Product Rule) f(x)g(x) differentiable at x₀ with derivative f'(x₀)g(x₀) + f(x₀)g'(x₀) (c) (Quotient Rule) If g(x₀) ≠ 0, f(x)/g(x) differentiable at x₀ with derivative (f'(x₀)g(x₀) - f(x₀)g'(x₀))/g(x₀)²

Question 32

# Proof Algebraic Properties of Differentiation Suppose f, g: E → ℝ are differentiable at x₀ (a) (Linearity) a, b constants a, b ∈ ℝ af(x) + bg(x) is differentiable at x₀ with derivative af'(x₀) + bg'(x₀) (b) (Product Rule) f(x)g(x) differentiable at x₀ with derivative f'(x₀)g(x₀) + f(x₀)g'(x₀) (c) (Quotient Rule) If g(x₀) ≠ 0, f(x)/g(x) differentiable at x₀ with derivative (f'(x₀)g(x₀) - f(x₀)g'(x₀))/g(x₀)² | 3 points

Answer 32

* (Linearity & Product Rule) Use f(x₀ + h) = f(x₀) + f'(x₀)h + ε₁(h)h, ε₁(h) → 0 * (Quotient Rule) Show d/dx 1/g(x) = -g'(x₀)/g(x₀)² * Apply product rule

Question 33

# Theorem Chain Rule

Answer 33

Assume f: E → ℝ, g: E' → ℝ f(E) ⊆ E' f is differentiable at x₀ ∈ E, which is a limit point of E g is differentiable at f(x₀) ∈ E' Then g(f(x)) is differentiable at x₀ with derivative g'(f(x₀))f'(x₀)

Question 34

# Proof (Chain Rule) Assume f: E → ℝ, g: E' → ℝ f(E) ⊆ E' f is differentiable at x₀ ∈ E, which is a limit point of E g is differentiable at f(x₀) ∈ E' Then g(f(x)) is differentiable at x₀ with derivative g'(f(x₀))f'(x₀) | 4 point

Answer 34

* f(x₀ + h) = f(x₀) + f'(x₀)h + ε₁(h) * g(f(x₀) + η) = g(f(x₀)) + g'(f(x₀))η + ε₂(η)η * Define ε₂(0) = 0 so ε₂ continuous at 0 * Show g(f(x₀ + h)) = g(f(x₀)) + g'(f(x₀))f'(x₀)h + [ ... ]h where [ ... ] → 0 as h → 0

Question 35

# Theorem Inverse Function Theorem IFT

Answer 35

Suppose I interval (non-trivial), f: I → ℝ strictly monotonic and continuous Let g: f(I) → ℝ be the (continuous, strictly monotonic) inverse Suppose x₀ ∈ I and f is differentiable at x₀ with f'(x₀) ≠ 0 Then g is differentiable at f(x₀) g'(f(x₀)) = 1/f'(x₀)

Question 36

# Proof (Inverse Function Theorem IFT) Suppose I interval (non-trivial), f: I → ℝ strictly monotonic and continuous Let g: f(I) → ℝ be the (continuous, strictly monotonic) inverse Suppose x₀ ∈ I and f is differentiable at x₀ with f'(x₀) ≠ 0 Then g is differentiable at f(x₀) g'(f(x₀)) = 1/f'(x₀) | 1 point

Answer 36

* Use derivative definition

Question 37

# Theorem Differentiation Theorem for Power Series

Answer 37

Suppose Σₙ₌₀ aₙxⁿ has ROC R > 0 (or R = ∞) Define f(x) = Σₙ₌₀ aₙxⁿ for |x| < R Then for |x| < R f is differentiable at x f'(x) = Σₙ₌₁ naₙxⁿ⁻¹

Question 38

# Theorem Fermat's Theorem on Extrema

Answer 38

Let f: (a, b) → ℝ, x₀ ∈ (a, b) is extremum of f Suppose f is differentiable at x₀ Then, f'(x₀) = 0

Question 39

# Proof (Fermat's Theorem on Extrema) Let f: (a, b) → ℝ, x₀ ∈ (a, b) is extremum of f Suppose f is differentiable at x₀ Then, f'(x₀) = 0 | 2 points

Answer 39

* Consider right and left derivatives * Show one is ≤ 0 and the other is ≥ 0

Question 40

# Theorem Rolle's Theorem

Answer 40

Suppose f: [a, b] → ℝ, such that (a) f is continuous on [a, b] (b) f is differentiable on [a, b] (c) f(a) = f(b) Then ∃ ξ ∈ (a, b): f'(ξ) = 0

Question 41

# Proof (Rolle's Theorem) Suppose f: [a, b] → ℝ, such that (a) f is continuous on [a, b] (b) f is differentiable on [a, b] (c) f(a) = f(b) Then ∃ ξ ∈ (a, b): f'(ξ) = 0 | 2 points

Answer 41

* Apply Boundedness Theorem * Apply Fermat's Theorem on Extrema

Question 42

# Theorem Mean Value Theorem

Answer 42

Suppose f: [a, b] → ℝ such that (a) f is countinuous on [a, b] (b) f is differentiable on (a, b) Then ∃ ξ ∈ (a, b): f'(ξ) = (f(b) - f(a))/(b - a)

Question 43

# Proof (Mean Value Theorem) Suppose f: [a, b] → ℝ such that (a) f is countinuous on [a, b] (b) f is differentiable on (a, b) Then ∃ ξ ∈ (a, b): f'(ξ) = (f(b) - f(a))/(b - a) | 2 points

Answer 43

* Define F(x) = f(x) - f(a) - ((f(b) - f(a))/(b - a))(x - a) * Apply Rolle's Theorem

Question 44

# Theorem Cauchy's MVT / Generalised MVT | 2 points

Answer 44

Let f, g: [a, b] → ℝ such that (a) f, g continuous on [a, b] (b) f, g differentiable on (a, b) Then ∃ ξ ∈ (a, b): f'(ξ)(g(b) - g(a)) = g'(ξ)(f(b) - f(a)) If g' ≠ 0 for all x ∈ (a, b) then g(b) ≠ g(a) and f'(ξ)/g'(ξ) = (f(b) - f(a))/(g(b) - g(a))

Question 45

# Proof (Cauchy's MVT / Generalised MVT) Let f, g: [a, b] → ℝ such that (a) f, g continuous on [a, b] (b) f, g differentiable on (a, b) Then ∃ ξ ∈ (a, b): f'(ξ)(g(b) - g(a)) = g'(ξ)(f(b) - f(a)) If g' ≠ 0 for all x ∈ (a, b) then g(b) ≠ g(a) and f'(ξ)/g'(ξ) = (f(b) - f(a))/(g(b) - g(a)) | 3 points

Answer 45

* For g(b) ≠ g(a) define F(x) = (f(x) - f(a)) - K(g(x) - g(a)) where K = (f(b) - f(a))/(g(b) - g(a)) * Apply Rolle's Theorem * For g(b) = g(a) apply Rolle's Theorem to g

Question 46

# Theorem Constancy Theorem

Answer 46

Suppose I is an interval and f: I → ℝ is such that f'(x) = 0 for all x ∈ I Then f is constant

Question 47

# Proof (Constancy Theorem) Suppose I is an interval and f: I → ℝ is such that f'(x) = 0 for all x ∈ I Then f is constant | 1 point

Answer 47

* Apply Mean Value Theorem to two members of I

Question 48

# Theorem Real Binomial Theorem

Answer 48

If -1 < x < 1, p ∈ ℝ (1 + x)ᵖ = Σ (ᵖₖ)xᵏ

Question 49

# Proof (Real Binomial Theorem) If -1 < x < 1, p ∈ ℝ (1 + x)ᵖ = Σ (ᵖₖ)xᵏ | 4 points

Answer 49

* Let g(x) = eᵖˡᵒᵍ⁽¹ ⁺ ˣ⁾, h(x) = Σ (ᵖₖ)xᵏ * g'(x) = (p/(1+x))g(x) and similarly for h * Define F(x) = h(x)/g(x) * Apply Constancy Theorem

Question 50

# Theorem Taylor's Theorem

Answer 50

Assume f: [a, b] → ℝ is such that f, f', f'', ..., f⁽ⁿ⁾ exist and are continuous on [a, b] and f⁽ⁿ⁺¹⁾ exists on (a, b) Then ∃ ξ ∈ (a, b) such that f(b) = f(a) + f'(a)(b - a) + (f''(a)/2)(b - a)² + ... + (f⁽ⁿ⁾(a)/n!)(b - a)ⁿ + (f⁽ⁿ⁺¹⁾(ξ)/(n + 1)!)(b - a)ⁿ⁺¹

Question 51

# Proof (Taylor's Theorem) Assume f: [a, b] → ℝ is such that f, f', f'', ..., f⁽ⁿ⁾ exist and are continuous on [a, b] and f⁽ⁿ⁺¹⁾ exists on (a, b) Then ∃ ξ ∈ (a, b) such that f(b) = f(a) + f'(a)(b - a) + (f''(a)/2)(b - a)² + ... + (f⁽ⁿ⁾(a)/n!)(b - a)ⁿ + (f⁽ⁿ⁺¹⁾(ξ)/(n + 1)!)(b - a)ⁿ⁺¹ | 7 point

Answer 51

* Use induction on n * n = 0 is MVT * Define F(x) = f(x) - f(a) - f'(a)(x - a) - ... - (f⁽ⁿ⁾(a)/n!)(x - a)ⁿ - (K/(n + 1)!)(x - a)ⁿ⁺¹ where K is s.t. F(b) = 0 * Apply Rolle's Theorem to get c s.t. F'(c) = 0 * Apply n - 1 case to f' on [a, c] * F'(c) = 0 substituting in f'(c) ⇒ K = f⁽ⁿ⁺¹⁾(ξ) * Subsitute x = b into F(x)

Question 52

# Theorem Simple L'Hôpital | 3 points

Answer 52

Suppose f, g: E → ℝ, p ∈ E limit point (a) f(p) = g(p) = 0 (b) f'(p), g'(p) exist (c) g'(p) ≠ 0 Then limₓ→ₚ f(x)/g(x) exists and = f'(p)/g'(p)

Question 53

# Proof (Simple L'Hôpital) Suppose f, g: E → ℝ, p ∈ E limit point (a) f(p) = g(p) = 0 (b) f'(p), g'(p) exist (c) g'(p) ≠ 0 Then limₓ→ₚ f(x)/g(x) exists and = f'(p)/g'(p) | 2 points

Answer 53

* Rearrange f(x)/g(x) * Use AOL

Question 54

# Theorem L'Hôpital 0/0 Form | 4 points

Answer 54

Suppose f, g real valued defined in some interval (p, p + δ) δ > 0 Also suppose: (a) f, g differentiable on (p, p + δ) (b) f(x), g(x) → 0 as x → p⁺ (c) g'(x) ≠ 0 on (p, p + δ) (d) limₓ→ₚ₊ f'(x)/g'(x) exisits (possibly ±∞) Then limₓ→ₚ₊ f(x)/g(x) = limₓ→ₚ₊ f'(x)/g'(x) Also g(x) ≠ 0 on some (p, p + δ') δ' > 0

Question 55

# Proof (L'Hôpital 0/0 Form) Suppose f, g real valued defined in some interval (p, p + δ) δ > 0 Also suppose: (a) f, g differentiable on (p, p + δ) (b) f(x), g(x) → 0 as x → p⁺ (c) g'(x) ≠ 0 on (p, p + δ) (d) limₓ→ₚ₊ f'(x)/g'(x) exisits (possibly ±∞) Then limₓ→ₚ₊ f(x)/g(x) = limₓ→ₚ₊ f'(x)/g'(x) Also g(x) ≠ 0 on some (p, p + δ') δ' > 0 | 4 points

Answer 55

* Define f(p) = g(p) = 0 * Fix x ∈ (p, p + δ) * Use Cauchy MVT * Let x → p⁺

Question 56

# Theorem L'Hôpital ∞/∞ Form | 4 points

Answer 56

If f, g: (a, a + δ) → ℝ (a) f, g differentiable on (a, a + δ) (b) |f|, |g| → ∞ (c) g' ≠ 0 on (a, a + δ) (d) f'(x)/g'(x) → L ∈ ℝ ∪ {±∞} as x → a⁺ Then f(x)/g(x) → L as x → a⁺ Also g ≠ 0 on (a, a + δ') for some δ' > 0

Question 57

# Proof (L'Hôpital ∞/∞ Form) If f, g: (a, a + δ) → ℝ (a) f, g differentiable on (a, a + δ) (b) |f|, |g| → ∞ (c) g' ≠ 0 on (a, a + δ) (d) f'(x)/g'(x) → L ∈ ℝ ∪ {±∞} as x → a⁺ Then f(x)/g(x) → L as x → a⁺ Also g ≠ 0 on (a, a + δ') for some δ' > 0 | 4 points

Answer 57

* Fix c ∈ (a, a + δ), x ∈ (a, c) * Use Cauchy MVT on [x, c] * |f'(ξ)/g'(ξ) - L| < ε for all ξ ∈ (a, c) * Rearrange to |f(x)/g(x) - L|