Suppose thatxn →x asn→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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→ ∞. Thenxnk →x ask → ∞ for all subsequences{xnk}. 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±xnwe 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±xnwe 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±xnwe 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±xnwe 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±xnwe 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{xn}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{xn}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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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 ≤xnk ≤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)