advanced calculus (i) - math.ncku.edu.tw

Advanced Calculus (I)

WEN-CHING LIEN

Department of MathematicsNational Cheng Kung University

WEN-CHING LIEN Advanced Calculus (I)

2.2 Limit Theorems

Theorem (Squeeze Theorem)

Suppose that {xn}, {yn}, and {wn} are real sequences.

(i) If xn → a and yn → a (the SAME a) as n → ∞, and ifthere is an N0 ∈ N such that

xn ≤ wn ≤ yn for n ≥ N0,

then wn → a as n → ∞.

(ii) If xn → 0 as n → ∞ and {yn} is bounded, thenxnyn → 0 as n → ∞.

2.2 Limit Theorems

Proof:

(i)Let ǫ > 0. Since xn and yn converge to a, use Definition2.1 and Theorem 1.6 to choose N1, N2 ∈ N such thatn ≥ N1 implies −ǫ ≤ xn − a ≤ ǫ and n ≥ N2 implies−ǫ ≤ yn − a ≤ ǫ. Set N = max{N0, N1, N2}. If n ≥ N wehave by hypothesis and the choice of N1 and N2 that

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

i.e., |wn − a| ≤ ǫ for n ≥ N. We conclude that wn → a asn → ∞.

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

Proof:

a − ǫ ≤ xn ≤ wn ≤ yn ≤ a + ǫ;

(ii)Suppose that xn → 0 and there is an M > 0 such that|yn| ≤ M for n ∈ N. Let ǫ > 0 and choose an N ∈ N suchthat n ≥ N implies |xn| ≤ ǫ/M. Then n ≥ N implies

|xnyn| ≤ Mǫ

M= ǫ.

We conclude that xnyn → 0 as n → ∞. 2

|xnyn| ≤ Mǫ

M= ǫ.

|xnyn| ≤ Mǫ

M= ǫ.

|xnyn| ≤ Mǫ

M= ǫ.

|xnyn| ≤ Mǫ

M= ǫ.

TheoremLet E ⊂ R. If E has a finite supermum (respectively, afinite infimum), then there is a sequence xn ∈ E such thatxn → sup E (respectively, xn → inf E) as n → ∞

Proof:

Suppose that E has a finite supermum. For each n ∈ N,choose (by the Approximation Property for Supermum) anxn ∈ E such that sup E − 1/n < xn ≤ sup E . Then by theSqueeze Theorem and Example 2.2, xn → sup E asn → ∞. Similarly, there is a sequence yn ∈ E such thatyn → inf E . 2

Proof:

TheoremSuppose that {xn} and {yn} are real sequences andα ∈ R. If {xn} and {yn} are convergent, then

(i)lim

n→∞

(xn + yn) = limn→∞

xn + limn→∞

(ii)lim

n→∞

(αxn) = α limn→∞

and(iii)

limn→∞

(xnyn) = ( limn→∞

xn)( limn→∞

(i)lim

n→∞

xn + limn→∞

(ii)lim

n→∞

and(iii)

limn→∞

xn)( limn→∞

(i)lim

n→∞

xn + limn→∞

(ii)lim

n→∞

and(iii)

limn→∞

xn)( limn→∞

(i)lim

n→∞

xn + limn→∞

(ii)lim

n→∞

and(iii)

limn→∞

xn)( limn→∞

(i)lim

n→∞

xn + limn→∞

(ii)lim

n→∞

and(iii)

limn→∞

xn)( limn→∞

TheoremIf, in addition, yn 6= 0 and limn→∞

yn 6= 0, then

limn→∞

limn→∞xn

limn→∞yn

(In particular, all these limits exists.)

TheoremIf, in addition, yn 6= 0 and limn→∞

yn 6= 0, then

limn→∞

limn→∞xn

limn→∞yn

(In particular, all these limits exists.)

DefinitionLet {xn} be a sequence of real numbers.

(i) {xn} is said to diverge to +∞ (notation: xn → +∞ asn → ∞ or limn→∞

xn = +∞) if and only if for each M ∈ Rthere is an N ∈ N such that

n ≥ N implies xn > M

(ii) {xn} is said to diverge to −∞ (notation: xn → −∞ asn → ∞ or limn→∞

xn = −∞) if and only if for each M ∈ Rthere is an N ∈ N such that

n ≥ N implies xn < M

TheoremSuppose that {xn} and {yn} are real sequences such thatxn → +∞ (respectively, xn → −∞) as n → ∞.

(i) If yn is bounded below (respectively, yn is boundedabove), then

limn→∞

(xn + yn) = +∞ (respectively , limn→∞

(xn + yn) = −∞).

(ii) If α > 0, then

limn→∞

(αxn) = +∞ (respectively , limn→∞

(αxn) = −∞).

limn→∞

(xn + yn) = −∞).

limn→∞

(αxn) = −∞).

limn→∞

(xn + yn) = −∞).

limn→∞

(αxn) = −∞).

limn→∞

(xn + yn) = −∞).

limn→∞

(αxn) = −∞).

Theorem(iii)If yn > M0 for some M0 > 0 and all n ∈ N, then

limn→∞

(xnyn) = +∞ (respectively , limn→∞

(xnyn) = −∞).

(iv) If {yn} is bounded and xn 6= 0, then

limn→∞

xn= 0.

limn→∞

(xnyn) = −∞).

limn→∞

xn= 0.

limn→∞

(xnyn) = −∞).

limn→∞

xn= 0.

Corollary:

Let {xn}, {yn} be real sequences and α, x, y be extendedreal numbers. If xn → x and yn → y , as n → ∞, then

limn→∞

(xn + yn) = x + y

(provided that the right side is not of the form ∞−∞), and

limn→∞

(αxn) = αx , limn→∞

(xnyn) = xy

(provided that none of these products is of the form0 · ±∞).

Corollary:

limn→∞

(xn + yn) = x + y

limn→∞

(xnyn) = xy

Corollary:

limn→∞

(xn + yn) = x + y

limn→∞

(xnyn) = xy

Corollary:

limn→∞

(xn + yn) = x + y

limn→∞

(xnyn) = xy

Theorem (Comparison Theorem)

Suppose that {xn} and {yn} are convergence sequences.If there is an N0 ∈ N such that

(1) xn ≤ yn n ≥ N0,

thenlim

n→∞

xn ≤ limn→∞

In particular, if xn ∈ [a, b] converges to some point c, thenc must belong to [a, b].

(1) xn ≤ yn n ≥ N0,

thenlim

n→∞

xn ≤ limn→∞

(1) xn ≤ yn n ≥ N0,

thenlim

n→∞

xn ≤ limn→∞

Proof:

Suppose that the first statement is false, i.e., that (1)holds but x := limn→∞

xn is great than y := limn→∞yn. Set

ǫ = (x − y)/2. Choose N1 > N0 such that |xn − x | < ǫ and|yn − y | < ǫ for n ≥ N1. Then for such an n,

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

which contradicts (1). This prove the first statement.

We conclude by noting that the second statement followsfrom the first, since a ≤ xn ≤ b implies a ≤ c ≤ b. 2

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Proof:

xn > x − ǫ = x −x − y

2= y +

x − y2

= y + ǫ > yn,

Thank you.

advanced calculus (i) - math.ncku.edu.tw

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