Advanced Calculus (I)

Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN Advanced Calculus (I)

2.5 Limits Superemum and Infimum

Definition

Let{x_n}ba a real sequence. Then thelimit supremumof {xn}is the extended real number

lim sup

n→∞

xn:= lim

n→∞(sup

k≥n

xk),

and thelimit infimum of{x_n}is the extended real number lim inf

n→∞ xn := lim

n→∞(inf

k≥nxk),

WEN-CHINGLIEN Advanced Calculus (I)

2.5 Limits Superemum and Infimum

Definition

Let{x_n}ba a real sequence. Then thelimit supremumof {xn}is the extended real number

lim sup

n→∞

xn:= lim

n→∞(sup

k≥n

xk),

and thelimit infimum of{x_n}is the extended real number lim inf

n→∞ xn := lim

n→∞(inf

k≥nxk),

WEN-CHINGLIEN Advanced Calculus (I)

2.5 Limits Superemum and Infimum

Definition

Let{x_n}ba a real sequence. Then thelimit supremumof {xn}is the extended real number

lim sup

n→∞

xn:= lim

n→∞(sup

k≥n

xk),

and thelimit infimum of{x_n}is the extended real number lim inf

n→∞ xn := lim

n→∞(inf

k≥nxk),

WEN-CHINGLIEN Advanced Calculus (I)

Theorem

Let{x_n}be a real sequence of numbers, s=lim sup

n→∞

xn, and t =lim inf

n→∞ xn. Then there are subsequences{x_nk}k∈N

and{xl_j}j∈N such that xn_k →s as k → ∞and xl_j →t as j → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Let{x_n}be a real sequence of numbers, s=lim sup

n→∞

xn, and t =lim inf

n→∞ xn. Then there are subsequences{x_nk}k∈N

and{xl_j}j∈N such that xn_k →s as k → ∞and xl_j →t as j → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

We will prove the result for the limit supremum. A similar argument establishes the result for the limit infimum.Let sn =sup_k≥nxk and observe thatsn↓s asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

We will prove the result for the limit supremum.A similar argument establishes the result for the limit infimum. Let sn =sup_k≥nxk and observe thatsn↓s asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

We will prove the result for the limit supremum. A similar argument establishes the result for the limit infimum.Let sn =sup_k≥nxk and observe thatsn↓s asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

We will prove the result for the limit supremum. A similar argument establishes the result for the limit infimum. Let sn =sup_k≥nxk and observe thatsn↓s asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞,there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞,there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞,there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞,there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner,we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner,we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence,it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence,it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

s=∞. Then by definitionsn=∞for alln ∈N. Since s1 =∞, there is ann1∈Nsuch thatxn1 >1. Since sn₁ =∞, there is ann2≥n1+1>n1such thatxn₂ >2.

Continuing in this manner, we can choose a subsequence {x_nk}such thatxnk >k for allk ∈N. Hence, it follows from the Squeeze Theorem (see Exersise 6, p. 44) that

xnk → ∞=s ask → ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=−∞. Sincesn ≥xn for alln∈N,it follows from the Squeeze Theorem thatxn→ −∞=sasn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 2.

s=−∞. Sincesn ≥xn for alln∈N, it follows from the Squeeze Theorem thatxn→ −∞=sasn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

s=−∞. Sincesn ≥xn for alln∈N,it follows from the Squeeze Theorem thatxn→ −∞=sasn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 2.

s=−∞. Sincesn ≥xn for alln∈N, it follows from the Squeeze Theorem thatxn→ −∞=sasn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 3.

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum),there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 3.

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum),there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly,there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 3.

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly,there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner,we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 3.

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner,we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞,we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 3.

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞,we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<s<∞. Setn0=0. By Theorem 1.20 (the Approximation Property for Supremum), there is an integern1 ∈Nsuch thatsn0+1−1<xn1 ≤sn0 +1.

Simiarly, there is an integern2 ≥n1+1>n1such that sn₁+1−1/2<xn₂ ≤sn₁+1. Continuing in this manner, we can choose integersn1<n2<· · · such that

snk−1+1− 1

k <xn_k ≤snk−1+1

fork ∈N. Sincesnk−1+1→s ask → ∞, we conclude by the Squeeze Theorem thatxnk →sask → ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Theorem

Let{xn}be a real sequence and x be an extended real number. Then xn →x as n→ ∞if and only if

(6) lim sup

n→∞

xn =lim inf

n→∞ xn =x.

WEN-CHINGLIEN Advanced Calculus (I)

Let{xn}be a real sequence and x be an extended real number. Then xn →x as n→ ∞if and only if

(6) lim sup

n→∞

xn =lim inf

n→∞ xn =x.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x; i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x;i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x; i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x;i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x; i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxn →x asn→ ∞. Thenxn_k →x ask → ∞ for all subsequences{x_nk}. Hence by Theorem 2.35, lim sup

n→∞

xn =x and lim inf

n→∞ xn=x; i.e., (6) holds.

Conversely, suppose that (6) holds.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

x =±∞. By considering±x_nwe may suppose that x =∞. Thus givenM ∈Rthere is anN ∈Nsuch that infk≥Nxk >M. It follows thatxn>M for all n≥N; i.e., xn → ∞asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

x =±∞. By considering±x_nwe may suppose that x =∞. Thus givenM ∈Rthere is anN ∈Nsuch that infk≥Nxk >M. It follows thatxn>M for all n≥N; i.e., xn → ∞asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

x =±∞. By considering±x_nwe may suppose that x =∞. Thus givenM ∈Rthere is anN ∈Nsuch that infk≥Nxk >M. It follows thatxn>M for all n≥N; i.e., xn → ∞asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

x =±∞. By considering±x_nwe may suppose that x =∞. Thus givenM ∈Rthere is anN ∈Nsuch that infk≥Nxk >M. It follows thatxn>M for all n≥N; i.e., xn → ∞asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

Case 1.

x =±∞. By considering±x_nwe may suppose that x =∞. Thus givenM ∈Rthere is anN ∈Nsuch that infk≥Nxk >M. It follows thatxn>M for all n≥N; i.e., xn → ∞asn→ ∞.

WEN-CHINGLIEN Advanced Calculus (I)

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< . We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 2.

−∞<x <∞. Let >0. ChooseN ∈Nsuch that

sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< . We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< . We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 2.

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< . We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< . We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Case 2.

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< .

We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

−∞<x <∞. Let >0. ChooseN ∈Nsuch that sup

k≥N

xk −x < and x− inf

k≥Nxk < .

Thus, for anyn≥N, xn−x ≤sup

k≥N

xk −x < and x−xn ≤x− inf

k≥Nxk < .

That is, for anyn≥N,

|x−xn|< .

We conclude thatxn →x asn→ ∞. 2

WEN-CHINGLIEN Advanced Calculus (I)

Theorem

Let{xn}be a real sequence of real numbers. Then lim sup

n→∞

xn (respectively,lim inf

n→∞ xn) is the largest value (respectively, the smallest value) to which some

subsequence of{x_n}converges. Namely, if xnk →x as k → ∞, then

(7) lim inf

n→∞ xn≤x ≤lim sup

n→∞

xn.

WEN-CHINGLIEN Advanced Calculus (I)

Let{xn}be a real sequence of real numbers. Then lim sup

n→∞

xn (respectively,lim inf

n→∞ xn) is the largest value (respectively, the smallest value) to which some

subsequence of{x_n}converges. Namely, if xnk →x as k → ∞, then

(7) lim inf

n→∞ xn≤x ≤lim sup

n→∞

xn.

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32,we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32,we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Proof:

Suppose thatxnk →x ask → ∞. FixN ∈Nand choose K so large thatk ≥K impliesnk ≥N. Clearly,

j≥Ninfxj ≤xn_k ≤sup

j≥N

xj

for allk ≥K. Taking the limit of this inequality ask → ∞, we obtain

j≥Ninfxj ≤x ≤sup

j≥N

xj

Taking the limit of this last inequality asN → ∞and applying Definition 2.32, we obtain(7). 2

WEN-CHINGLIEN Advanced Calculus (I)

Remark:

If{x_n}is any sequence of real numbers, then lim inf

n→∞ xn ≤lim sup

n→∞

xn.

WEN-CHINGLIEN Advanced Calculus (I)

Remark:

If{x_n}is any sequence of real numbers, then lim inf

n→∞ xn ≤lim sup

n→∞

xn.

WEN-CHINGLIEN Advanced Calculus (I)

Thank you.

WEN-CHINGLIEN Advanced Calculus (I)