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Prelims Mathematics and Computer Science > M2: Analysis II > Flashcards

M2: Analysis II Flashcards

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1
Q

Proposition

Limit Points via Sequences

A

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

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2
Q

Proof

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

2 points

A
  • (⇒) Sandwich with δ = 1/n
  • (⇐) Use convergence definition
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3
Q

Proposition

Limits via Sequences

A

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 → ∞.

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4
Q

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

A
  • (⇒) Use δ in limit definition as ε in convergence definition
  • (⇐) Use contrapositive
  • Let δ = 1/n to find sequence
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5
Q

Theorem

AOL for Functions

4 points

A

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

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6
Q

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

A
  • Pick pₙ ∈ E\{p} pₙ → p
  • Use Analysis I AOL
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7
Q

Theorem

Extended AOL

4 points

A

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

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8
Q

Theorem

Limits Preserve Weak Inequalities

A

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

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8
Q

Theorem

Sandwiching

A

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

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8
Q

Theorem

Algebra of Continuous Functions

A

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

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9
Q

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

A
  • Immediate for p isolated
  • Use AOL for p limit point
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9
Q

Theorem

Continuous Functions Commute with Limits

A

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)

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10
Q

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

A
  • Use continuiuty and limit definitions
  • Use δ in first definition as ε in second
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11
Q

Theorem

Composition of Continuous Functions

A

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

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12
Q

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

A
  • Immediate for p isolated
  • Use continuous functions commute with limits for p limit point
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13
Q

Theorem

Boundedness Theorem

A

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

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14
Q

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

A
  • 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
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15
Q

Theorem

Intermediate Value Theorem IVT

A

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

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16
Q

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

A
  • 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
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17
Q

Theorem

Continuous Inverse Function Theorem CIFT

A

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

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18
Q

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

A
  • 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)
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19
Q

Proposition

Uniform Continuity via Sequences

A

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

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20
Q

Proof

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

4 points

A
  • (⇒) 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
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21
Q

Theorem

Continuity Implies Uniform Continuity on Closed Bounded Intervals

A

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

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22
Q

Proof

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

4 points

A
  • 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
23
Q

Theorem

Uniform Limits Preserve Continuity

A

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

24
Q

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

A
  • |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
25
Q

Theorem

Cauchy Convergence Criterion for Uniform Convergence of Sequences

A

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

26
Q

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

A
  • (⇒) 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
27
Q

Theorem

Weierstrass M-Test

A

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

28
Q

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

A
  • Use Cauchy convergence definition
29
Q

Theorem

Uniform Convergence and Continuity of Power Series

A

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

30
Q

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

A
  • Apply M-Test with Mₖ = |aₖ|ρᵏ
31
Q

Theorem

Algebraic Properties of Differentiation

3 points

A

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

32
Q

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

A
  • (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
33
Q

Theorem

Chain Rule

A

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)

34
Q

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

A
  • 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
35
Q

Theorem

Inverse Function Theorem IFT

A

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)

36
Q

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

A
  • Use derivative definition
37
Q

Theorem

Differentiation Theorem for Power Series

A

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

38
Q

Theorem

Fermat’s Theorem on Extrema

A

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

39
Q

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

A
  • Consider right and left derivatives
  • Show one is ≤ 0 and the other is ≥ 0
40
Q

Theorem

Rolle’s Theorem

A

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

41
Q

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

A
  • Apply Boundedness Theorem
  • Apply Fermat’s Theorem on Extrema
42
Q

Theorem

Mean Value Theorem

A

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)

43
Q

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

A
  • Define F(x) = f(x) - f(a) - ((f(b) - f(a))/(b - a))(x - a)
  • Apply Rolle’s Theorem
44
Q

Theorem

Cauchy’s MVT / Generalised MVT

2 points

A

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))

45
Q

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

A
  • 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
46
Q

Theorem

Constancy Theorem

A

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

47
Q

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

A
  • Apply Mean Value Theorem to two members of I
48
Q

Theorem

Real Binomial Theorem

A

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

49
Q

Proof

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

4 points

A
  • 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
50
Q

Theorem

Taylor’s Theorem

A

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

51
Q

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

A
  • 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)
52
Q

Theorem

Simple L’Hôpital

3 points

A

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)

53
Q

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

A
  • Rearrange f(x)/g(x)
  • Use AOL
54
Q

Theorem

L’Hôpital 0/0 Form

4 points

A

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

55
Q

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

A
  • Define f(p) = g(p) = 0
  • Fix x ∈ (p, p + δ)
  • Use Cauchy MVT
  • Let x → p⁺
56
Q

Theorem

L’Hôpital ∞/∞ Form

4 points

A

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

57
Q

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

A
  • 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|

Prelims Mathematics and Computer Science (7 decks)

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