Advanced Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
WEN-CHINGLIEN 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 if there 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, then xnyn →0 as n→ ∞.
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 if there 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, then xnyn →0 as n → ∞.
WEN-CHINGLIEN 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 if there 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, then xnyn →0 as n→ ∞.
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 if there 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, then xnyn →0 as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a,use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a,use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
(i)
Letǫ >0. Since xn and ynconverge to a, use Definition 2.1 and Theorem 1.6 to choose N1,N2∈N such that n ≥N1implies−ǫ ≤xn−a≤ ǫand n≥N2 implies
−ǫ ≤yn−a≤ ǫ. Set N =max{N0,N1,N2}. If n≥N we have by hypothesis and the choice of N1and N2that
a− ǫ ≤xn ≤wn≤yn ≤a+ ǫ;
i.e.,|wn−a| ≤ ǫfor n≥N. We conclude that wn→a as n → ∞.
(ii)
Suppose that xn →0 and there is an M>0 such that
|yn| ≤M for n∈N. Letǫ >0 and choose an N ∈N such that n ≥N implies|xn| ≤ ǫ/M. Then n ≥N implies
|xnyn| ≤M ǫ M = ǫ.
We conclude that xnyn →0 as n→ ∞. 2
WEN-CHINGLIEN Advanced Calculus (I)
(ii)
Suppose that xn →0 and there is an M>0 such that
|yn| ≤M for n∈N. Letǫ >0 and choose an N ∈N such that n ≥N implies|xn| ≤ ǫ/M. Then n ≥N implies
|xnyn| ≤M ǫ M = ǫ.
We conclude that xnyn →0 as n→ ∞. 2
(ii)
Suppose that xn →0 and there is an M>0 such that
|yn| ≤M for n∈N. Letǫ >0 and choose an N ∈N such that n ≥N implies|xn| ≤ ǫ/M. Then n ≥N implies
|xnyn| ≤M ǫ M = ǫ.
We conclude that xnyn →0 as n→ ∞. 2
WEN-CHINGLIEN Advanced Calculus (I)
(ii)
Suppose that xn →0 and there is an M>0 such that
|yn| ≤M for n∈N. Letǫ >0 and choose an N ∈N such that n ≥N implies|xn| ≤ ǫ/M. Then n ≥N implies
|xnyn| ≤M ǫ M = ǫ.
We conclude that xnyn →0 as n→ ∞. 2
(ii)
Suppose that xn →0 and there is an M>0 such that
|yn| ≤M for n∈N. Letǫ >0 and choose an N ∈N such that n ≥N implies|xn| ≤ ǫ/M. Then n ≥N implies
|xnyn| ≤M ǫ M = ǫ.
We conclude that xnyn →0 as n→ ∞. 2
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Let E ⊂R. If E has a finite supermum (respectively, a finite infimum), then there is a sequence xn ∈E such that xn →sup E (respectively, xn→inf E) as n → ∞
Theorem
Let E ⊂R. If E has a finite supermum (respectively, a finite infimum), then there is a sequence xn ∈E such that xn →sup E (respectively, xn→inf E) as n → ∞
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly, there is a sequence yn ∈E such that yn →inf E. 2
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly, there is a sequence yn ∈E such that yn →inf E. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly, there is a sequence yn ∈E such that yn →inf E. 2
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly,there is a sequence yn ∈E such that yn →inf E. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly, there is a sequence yn ∈E such that yn →inf E. 2
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly,there is a sequence yn ∈E such that yn →inf E. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
Suppose that E has a finite supermum. For each n∈N, choose (by the Approximation Property for Supermum) an xn ∈E such that sup E −1/n<xn≤sup E. Then by the Squeeze Theorem and Example 2.2, xn →sup E as n → ∞. Similarly, there is a sequence yn ∈E such that yn →inf E. 2
Theorem
Suppose that{xn}and{yn}are real sequences and α ∈ R. If{xn}and{yn}are convergent, then
(i)
nlim→∞
(xn+yn) = lim
n→∞
xn+ lim
n→∞
yn, (ii)
nlim→∞
(αxn) = α lim
n→∞
xn
and (iii)
nlim→∞
(xnyn) = (lim
n→∞
xn)(lim
n→∞
yn).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that{xn}and{yn}are real sequences and α ∈ R. If{xn}and{yn}are convergent, then
(i)
nlim→∞
(xn+yn) = lim
n→∞
xn+ lim
n→∞
yn, (ii)
nlim→∞
(αxn) = α lim
n→∞
xn
and (iii)
nlim→∞
(xnyn) = (lim
n→∞
xn)(lim
n→∞
yn).
Theorem
Suppose that{xn}and{yn}are real sequences and α ∈ R. If{xn}and{yn}are convergent, then
(i)
nlim→∞
(xn+yn) = lim
n→∞
xn+ lim
n→∞
yn, (ii)
nlim→∞
(αxn) = α lim
n→∞
xn
and (iii)
nlim→∞
(xnyn) = (lim
n→∞
xn)(lim
n→∞
yn).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that{xn}and{yn}are real sequences and α ∈ R. If{xn}and{yn}are convergent, then
(i)
nlim→∞
(xn+yn) = lim
n→∞
xn+ lim
n→∞
yn, (ii)
nlim→∞
(αxn) = α lim
n→∞
xn
and (iii)
nlim→∞
(xnyn) = (lim
n→∞
xn)(lim
n→∞
yn).
Theorem
Suppose that{xn}and{yn}are real sequences and α ∈ R. If{xn}and{yn}are convergent, then
(i)
nlim→∞
(xn+yn) = lim
n→∞
xn+ lim
n→∞
yn, (ii)
nlim→∞
(αxn) = α lim
n→∞
xn
and (iii)
nlim→∞
(xnyn) = (lim
n→∞
xn)(lim
n→∞
yn).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
If, in addition, yn6=0 and limn→∞yn6=0, then (iv)
nlim→∞
xn
yn
= limn→∞xn
limn→∞yn
(In particular, all these limits exists.)
Theorem
If, in addition, yn6=0 and limn→∞yn6=0, then (iv)
nlim→∞
xn
yn
= limn→∞xn
limn→∞yn
(In particular, all these limits exists.)
WEN-CHINGLIEN Advanced Calculus (I)
Definition
Let{xn}be a sequence of real numbers.
(i){xn}is said to diverge to+∞(notation: xn → +∞as n → ∞or limn→∞xn = +∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn >M
(ii){xn}is said to diverge to−∞(notation: xn→ −∞as n → ∞or limn→∞xn = −∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn <M
Definition
Let{xn}be a sequence of real numbers.
(i){xn}is said to diverge to+∞(notation: xn → +∞as n → ∞or limn→∞xn = +∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn >M
(ii){xn}is said to diverge to−∞(notation: xn→ −∞as n → ∞or limn→∞xn = −∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn <M
WEN-CHINGLIEN Advanced Calculus (I)
Definition
Let{xn}be a sequence of real numbers.
(i){xn}is said to diverge to+∞(notation: xn → +∞as n → ∞or limn→∞xn = +∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn >M
(ii){xn}is said to diverge to−∞(notation: xn→ −∞as n → ∞or limn→∞xn = −∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn <M
Definition
Let{xn}be a sequence of real numbers.
(i){xn}is said to diverge to+∞(notation: xn → +∞as n → ∞or limn→∞xn = +∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn >M
(ii){xn}is said to diverge to−∞(notation: xn→ −∞as n → ∞or limn→∞xn = −∞) if and only if for each M ∈R there is an N ∈N such that
n ≥N implies xn <M
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that{xn}and{yn}are real sequences such that xn → +∞(respectively, xn→ −∞) as n→ ∞.
(i) If yn is bounded below (respectively, ynis bounded above), then
nlim→∞
(xn+yn) = +∞ (respectively, lim
n→∞
(xn+yn) = −∞).
(ii) Ifα >0, then
nlim→∞
(αxn) = +∞ (respectively, lim
n→∞
(αxn) = −∞).
Theorem
Suppose that{xn}and{yn}are real sequences such that xn → +∞(respectively, xn→ −∞) as n→ ∞.
(i) If yn is bounded below (respectively, ynis bounded above), then
nlim→∞
(xn+yn) = +∞ (respectively, lim
n→∞
(xn+yn) = −∞).
(ii) Ifα >0, then
nlim→∞
(αxn) = +∞ (respectively, lim
n→∞
(αxn) = −∞).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that{xn}and{yn}are real sequences such that xn → +∞(respectively, xn→ −∞) as n→ ∞.
(i) If yn is bounded below (respectively, ynis bounded above), then
nlim→∞
(xn+yn) = +∞ (respectively, lim
n→∞
(xn+yn) = −∞).
(ii) Ifα >0, then
nlim→∞
(αxn) = +∞ (respectively, lim
n→∞
(αxn) = −∞).
Theorem
Suppose that{xn}and{yn}are real sequences such that xn → +∞(respectively, xn→ −∞) as n→ ∞.
(i) If yn is bounded below (respectively, ynis bounded above), then
nlim→∞
(xn+yn) = +∞ (respectively, lim
n→∞
(xn+yn) = −∞).
(ii) Ifα >0, then
nlim→∞
(αxn) = +∞ (respectively, lim
n→∞
(αxn) = −∞).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
(iii)If yn >M0 for some M0>0 and all n∈N, then
nlim→∞
(xnyn) = +∞ (respectively, lim
n→∞
(xnyn) = −∞).
(iv) If{yn}is bounded and xn6=0, then
nlim→∞
yn
xn
=0.
Theorem
(iii)If yn >M0 for some M0>0 and all n∈N, then
nlim→∞
(xnyn) = +∞ (respectively, lim
n→∞
(xnyn) = −∞).
(iv) If{yn}is bounded and xn6=0, then
nlim→∞
yn
xn
=0.
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
(iii)If yn >M0 for some M0>0 and all n∈N, then
nlim→∞
(xnyn) = +∞ (respectively, lim
n→∞
(xnyn) = −∞).
(iv) If{yn}is bounded and xn6=0, then
nlim→∞
yn
xn
=0.
Corollary:
Let{xn}, {yn}be real sequences and α, x, y be extended real numbers. If xn→x and yn→y , as n→ ∞, then
nlim→∞
(xn+yn) =x+y
(provided that the right side is not of the form∞ − ∞), and
nlim→∞
(αxn) = αx, lim
n→∞
(xnyn) =xy (provided that none of these products is of the form 0· ±∞).
WEN-CHINGLIEN Advanced Calculus (I)
Corollary:
Let{xn}, {yn}be real sequences and α, x, y be extended real numbers. If xn→x and yn→y , as n→ ∞, then
nlim→∞
(xn+yn) =x+y
(provided that the right side is not of the form∞ − ∞), and
nlim→∞
(αxn) = αx, lim
n→∞
(xnyn) =xy (provided that none of these products is of the form 0· ±∞).
Corollary:
Let{xn}, {yn}be real sequences and α, x, y be extended real numbers. If xn→x and yn→y , as n→ ∞, then
nlim→∞
(xn+yn) =x+y
(provided that the right side is not of the form∞ − ∞), and
nlim→∞
(αxn) = αx, lim
n→∞
(xnyn) =xy (provided that none of these products is of the form 0· ±∞).
WEN-CHINGLIEN Advanced Calculus (I)
Corollary:
Let{xn}, {yn}be real sequences and α, x, y be extended real numbers. If xn→x and yn→y , as n→ ∞, then
nlim→∞
(xn+yn) =x+y
(provided that the right side is not of the form∞ − ∞), and
nlim→∞
(αxn) = αx, lim
n→∞
(xnyn) =xy (provided that none of these products is of the form 0· ±∞).
Theorem (Comparison Theorem)
Suppose that{xn}and{yn}are convergence sequences.
If there is an N0 ∈N such that
(1) xn≤yn n≥N0,
then
nlim→∞
xn ≤ lim
n→∞
yn.
In particular, if xn∈ [a,b]converges to some point c, then c must belong to [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (Comparison Theorem)
Suppose that{xn}and{yn}are convergence sequences.
If there is an N0 ∈N such that
(1) xn≤yn n≥N0,
then
nlim→∞
xn ≤ lim
n→∞
yn.
In particular, if xn∈ [a,b]converges to some point c, then c must belong to [a,b].
Theorem (Comparison Theorem)
Suppose that{xn}and{yn}are convergence sequences.
If there is an N0 ∈N such that
(1) xn≤yn n≥N0,
then
nlim→∞
xn ≤ lim
n→∞
yn.
In particular, if xn∈ [a,b]converges to some point c, then c must belong to [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n,
xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n,
xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
WEN-CHINGLIEN Advanced Calculus (I)
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
WEN-CHINGLIEN Advanced Calculus (I)
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first,since a≤xn ≤b implies a ≤c ≤b. 2
WEN-CHINGLIEN Advanced Calculus (I)
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first,since a≤xn ≤b implies a ≤c ≤b. 2
WEN-CHINGLIEN Advanced Calculus (I)
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>N0such that|xn−x| < ǫand
|yn−y| < ǫfor n ≥N1. Then for such an n, xn >x − ǫ =x− x −y
2 =y+ x−y
2 =y+ ǫ > yn, which contradicts (1). This prove the first statement.
We conclude by noting that the second statement follows from the first, since a ≤xn ≤b implies a ≤c ≤b. 2
Thank you.
WEN-CHINGLIEN Advanced Calculus (I)
