advanced calculus (i) - math.ncku.edu.tw
TRANSCRIPT
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Advanced Calculus (I)
WEN-CHING LIEN
Department of MathematicsNational Cheng Kung University
WEN-CHING LIEN Advanced Calculus (I)
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
WEN-CHING LIEN Advanced Calculus (I)
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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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 → ∞.
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(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
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(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
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(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
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(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
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(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
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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 → ∞
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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 → ∞
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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
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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
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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
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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
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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
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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
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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
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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→∞
yn,
(ii)lim
n→∞
(αxn) = α limn→∞
xn
and(iii)
limn→∞
(xnyn) = ( limn→∞
xn)( limn→∞
yn).
WEN-CHING LIEN Advanced Calculus (I)
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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→∞
yn,
(ii)lim
n→∞
(αxn) = α limn→∞
xn
and(iii)
limn→∞
(xnyn) = ( limn→∞
xn)( limn→∞
yn).
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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→∞
yn,
(ii)lim
n→∞
(αxn) = α limn→∞
xn
and(iii)
limn→∞
(xnyn) = ( limn→∞
xn)( limn→∞
yn).
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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→∞
yn,
(ii)lim
n→∞
(αxn) = α limn→∞
xn
and(iii)
limn→∞
(xnyn) = ( limn→∞
xn)( limn→∞
yn).
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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→∞
yn,
(ii)lim
n→∞
(αxn) = α limn→∞
xn
and(iii)
limn→∞
(xnyn) = ( limn→∞
xn)( limn→∞
yn).
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TheoremIf, in addition, yn 6= 0 and limn→∞
yn 6= 0, then
(iv)
limn→∞
xn
yn=
limn→∞xn
limn→∞yn
(In particular, all these limits exists.)
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TheoremIf, in addition, yn 6= 0 and limn→∞
yn 6= 0, then
(iv)
limn→∞
xn
yn=
limn→∞xn
limn→∞yn
(In particular, all these limits exists.)
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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
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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
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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
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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
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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) = −∞).
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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) = −∞).
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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) = −∞).
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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) = −∞).
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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→∞
yn
xn= 0.
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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→∞
yn
xn= 0.
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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→∞
yn
xn= 0.
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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 · ±∞).
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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 · ±∞).
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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 · ±∞).
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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 · ±∞).
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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→∞
yn.
In particular, if xn ∈ [a, b] converges to some point c, thenc must belong to [a, b].
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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→∞
yn.
In particular, if xn ∈ [a, b] converges to some point c, thenc must belong to [a, b].
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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→∞
yn.
In particular, if xn ∈ [a, b] converges to some point c, thenc must belong to [a, b].
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Thank you.
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