Chapter 11: Sequences; Indeterminate Forms, Improper Integrals
Transcript of Chapter 11: Sequences; Indeterminate Forms, Improper Integrals
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Section 11.1 The Least Upper Bound Axioma. Least Upper Bound Axiomb. Examplesc. Theorem 11.1.2d. Examplee. Greatest Lower Boundf. Theorem 11.1.4
Section 11.2 Sequences of Real Numbersa. Definitionb. Laws of Formationc. Types of Sequencesd. Range of a Sequencee. Increasing, Decreasing, Monotonic Sequencesf. Example
Section 11.3 Limit of A Sequencea. Definitionb. Examplec. Uniqueness of Limit Theorem d. Convergent and Divergent Sequencese. Convergence Theoremf. Exampleg. Theorem 11.3.7h. Pinching Theorem for Sequencesi. Corollaryj. Continuous Functions Applied to Convergent Sequences
Chapter 11: Sequences; Indeterminate Forms; Improper IntegralsSection 11.4 Some Important Limitsa. Common Limitsb. Limit Propertiesc. More Properties
Section 11.5 The Indeterminate Form (0/0)a. L’Hôpitals Rule (0/0)b. Examplec. The Cauchy Mean-Value Theorem
Section 11.6 The Indeterminate Form (∞/∞); Other Indeterminate Forms
a. L’Hôpitals Rule (∞/∞)b. Limit Propertiesc. Exampled. Indeterminates 00, 1∞, ∞0
e. Example
Section 11.7 Improper Integralsa. Integrals Over Unbounded Intervalsb. Examplesc. Limit Propertyd. Comparison Teste. Integrals of Unbounded Functionsf. Example
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The Least Upper Bound Axiom
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The Least Upper Bound Axiom
We indicate the least upper bound of a set S by writing lub S. As you will see from the examples below, the least upper bound idea has wide applicability.
(1) lub (−∞, 0) = 0, lub(−∞, 0] = 0.
(2) lub (−4,−1) = −1, lub(−4,−1] = −1.
(3) lub {1/2, 2/3, 3/4, . . . , n/(n + 1), . . . } = 1.
(4) lub {−1/2,−1/8,−1/27, . . . ,−1/n3, . . . } = 0.
(5) lub {x : x2 < 3} = lub{x : } =
(6) For each decimal fraction
b = 0.b1b2b3, . . . ,
we have
b = lub {0.b1, 0.b1b2, 0.b1b2b3, . . . }.
(7) If S consists of the lengths of all polygonal paths inscribed in a semicircle of radius 1, then lub S = π (half the circumference of the unit circle).
3 3x− < < 3
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The Least Upper Bound Axiom
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The Least Upper Bound Axiom
Example
(a) Let S = {1/2, 2/3, 3/4, . . . , n/(n + 1), . . . } and take ε = 0.0001. Since 1 is the least upper bound of S, there must be a number s ∈ S such that
1 − 0.0001 < s < 1.
There is: take, for example,
(b) Let S = {0, 1, 2, 3} and take ε = 0.00001. It is clear that 3 is the least upper bound of S. Therefore, there must be a number s ∈ S such that
3 − 0.00001 < s ≤ 3.There is: s = 3.
99,999 .100,000
s =
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The Least Upper Bound Axiom
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The Least Upper Bound Axiom
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Sequences of Real Numbers
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Sequences of Real NumbersSuppose we have a sequence of real numbers
a1, a2, a3, . . . , an, . . .
By convention, a1 is called the first term of the sequence, a2 the second term, and so on. More generally, an, the term with index n, is called the nth term.Sequences can be defined by giving the law of formation. For example:
2
1 1 12 3 4
1 2 3 42 3 4 5
1 is the sequence 1, , , , . . .
is the sequence , , , , . . .1
is the sequence 1, 4, 9, 16, . . .
n
n
n
an
nbn
c n
=
=+=
It’s like defining f by giving f (x).
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Sequences of Real Numbers
Sequences can be multiplied by constants; they can be added, subtracted, andmultiplied. From
a1, a2, a3, . . . , an, . . . and b1, b2, b3, . . . , bn, . . .
we can form
the scalar product sequence : αa1, αa2, αa3, . . . , αan, . . . ,
the sum sequence : a1 + b1, a2 + b2, a3 + b3, . . . , an + bn, . . . ,
the difference sequence : a1 − b1, a2 − b2, a3 − b3, . . . , an − bn, . . . ,
the product sequence : a1b1, a2b2, a3b3, . . . , anbn, . . . .
If the bi ’s are all different from zero, we can form
the quotient sequence :
the reciprocal sequence :1 2 3
1 1 1 1, , , . . . , , . . .nb b b b
31 2
1 2 3
, , , . . . , , . . .n
n
a aa ab b b b
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Sequences of Real Numbers
The range of a sequence is the set of values taken on by the sequence. While there can be repetition in a sequence, there can be no repetition in the statement of its range. The range is a set, and in a set there is no repetition. A number is either in a particular set or it’s not. It can’t be there more than once. The sequences
0, 1, 0, 1, 0, 1, 0, 1, . . . and 0, 0, 1, 1, 0, 0, 1, 1, . . .
both have the same range: the set {0, 1}. The range of the sequence0, 1,−1, 2, 2,−2, 3, 3, 3,−3, 4, 4, 4, 4,−4, . . .
is the set of integers.
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Sequences of Real NumbersMany of the sequences we work with have some regularity. They either have an upward tendency or they have a downward tendency. The following terminology is standard. The sequence with terms an is said to be
increasing if an < an+1 for all n,
nondecreasing if an ≤ an+1 for all n,
decreasing if an > an+1 for all n,
nonincreasing if an ≥ an+1 for all n.
A sequence that satisfies any of these conditions is called monotonic.The sequences
1, ½, 1∕3, ¼, . . . , 1∕n , . . .
2, 4, 8, 16, . . . , 2n, . . .
2, 2, 4, 4, 6, 6, . . . , 2n, 2n, . . .
are monotonic. The sequence
1, ½, 1, 1∕3, 1, ¼, 1, . . .
is not monotonic.
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Sequences of Real NumbersExampleThe sequence
is increasing. It is bounded below by ½ (the greatest lower bound) and above by 1 (the least upper bound).ProofSince
we have an < an+1. This confirms that the sequence is increasing. The sequence can be displayed as
It is clear that ½ is the greatest lower bound and 1 is the least upper bound.
1nna
n=
+
( ) ( )( )
21
2
1 / 2 1 1 2 1 1/ 1 2 2
n
n
n na n n n na n n n n n n+ + + + + + += = ⋅ = >
+ + +
1 2 3 4 98 99, , , ,. . . , , , . . .2 3 4 5 99 100
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Limit of A Sequence
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Limit of A SequenceExampleSince
it is intuitively clear that
To verify that this statement conforms to the definition of limit of a sequence, we must show that for each ε > 0 there exists a positive integer K such thatif n ≥ K, then
To do this, we fix ε > 0 and note that
We now choose K sufficiently large that . If n ≥ K, thenand consequently
2 21 1 1
nn n
=+ +
2lim 21n
nn→∞
=+
2 21
nn
ε− <+
( )2 2 121 1
2 2 221 1
n nnn n n n n
− +
+ +
−− = = = <
+ +
2 221
nn n
ε− < <+
2 K ε< 2 2n K ε≤ <
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Limit of A Sequence
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Limit of A Sequence
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Limit of A Sequence
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Limit of A Sequence
ExampleWe shall show that the sequence is convergent. Since
3 = (3n)1/n < (3n + 4n)1/n < (4n + 4n)1/n = (2 · 4n)1/n = 21/n · 4 ≤ 8,
the sequence is bounded. Note that
(3n + 4n)(n+1)/n = (3n + 4n)1/n(3n + 4n)= (3n + 4n)1/n3n + (3n + 4n)1/n4n.
Since(3n + 4n)1/n > (3n)1/n = 3 and (3n + 4n)1/n > (4n)1/n = 4,
we have(3n + 4n)(n+1)/n > 3 · (3n) + 4 · (4n) = 3n+1 + 4n+1.
Taking the (n + 1)st root of the left and right sides of this inequality, we obtain
(3n + 4n)1/n > (3n+1 + 4n+1)1/(n+1)
The sequence is decreasing. Being also bounded, it must be convergent.
( )1/3 4
nn nna = +
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Limit of A Sequence
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Limit of A Sequence
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Limit of A Sequence
The following is an obvious corollary to the pinching theorem.
A limit to remember
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Limit of A Sequence
Continuous Functions Applied to Convergent Sequences
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Some Important Limits
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Some Important Limits
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Some Important Limits
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The Indeterminate Form (0/0)
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The Indeterminate Form (0/0)
ExampleFind
SolutionAs x → 0+, both numerator and denominator tend to 0 and
It follows from L’Hôpital’s rule that
For short, we can write
0lim
sinx
xx+→
( )( ) ( )( )
1 2 0 01coscos 1/ 2
f x xg x xx x
′= = → =
′
0lim 0
sinx
xx+→→
0 0
2lim lim 0sin cosx x
x xx x+ +→ →
=
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The Indeterminate Form (0/0)
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Other Indeterminate Forms
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Other Indeterminate Forms
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Other Indeterminate Forms
ExampleDetermine the behavior of as n→∞.
SolutionTo use the methods of calculus, we investigate
Since both numerator and denominator tend to ∞ with x, we try L’Hôpital’s rule:
Therefore the sequence must also diverge to ∞.
2
2n
nan
=
2
2limx
x x→∞
( )2
2
2 ln 22 2 ln 2lim lim lim2 2
xx x
x x xx x→∞ →∞ →∞= ∞
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Other Indeterminate Forms
The Indeterminates 00, 1∞, ∞0
Such indeterminates are usually handled by first applying the logarithm function:
y = [ f (x)]g(x) gives ln y = g(x) ln f (x).
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Other Indeterminate FormsExample (1∞) Find
SolutionHere we are dealing with an indeterminate of the form 1∞: as x → 0+, 1 + x → 1 and 1/x →∞. Taking the logarithm and then applying L’Hôpital’s rule, we have
As x → 0+, ln (1 + x)1/x → 1 and (1 + x)1/x = eln(1+x)1/x → e1 = e. Set x = 1/nand we have the familiar result: as n→∞, [1 + (1/n)]n → e.
( )1/
0lim 1 x
xx
+→+
( ) ( )1/
0 0 0
ln 1 1lim ln 1 lim lim 11
x
x x x
xx
x x+ + +→ → →
++ = =
+
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Improper IntegralsIntegrals over Unbounded IntervalsWe begin with a function f which is continuous on an unbounded interval [a,∞). For each number b > a we can form the definite integral
If, as b tends to ∞, this integral tends to a finite limit L,
( )b
af x dx∫
( )limb
abf x dx L
→∞=∫
then we write
and say that
the improper integral
Otherwise, we say that
the improper integral
converges to L.
diverges.
( )a
f x dx L∞
=∫
( )a
f x dx∞
∫
( )a
f x dx∞
∫
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Improper Integrals
As a tends to −∞, sin πa oscillates between −1 and 1. Therefore the integral oscillates between 1/π and −1/π and does not converge.
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Improper Integrals
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Improper Integrals
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Improper IntegralsIntegrals of Unbounded FunctionsImproper integrals can arise on bounded intervals. Suppose that f is continuous on the half-open interval [a, b) but is unbounded there. For each number c < b, we can form the definite integral
If, as c → b−, the integral tends to a finite limit L, namely, if
( )c
af x dx∫
( )limc
ac bf x dx L
−→=∫
then we write
and say that
the improper integral
Otherwise, we say that the improper integral diverges.
converges to L.
( )b
af x dx L=∫
( )b
af x dx∫
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Improper IntegralsExampleTest(∗)
for convergence.
SolutionThe integrand has an infinite discontinuity at x = 2.For integral (∗) to converge both
must converge. Neither does. For instance, as c → 2−,
This tells us that
If we overlook the infinite discontinuity at x = 2, we can be led to the incorrect conclusion that
( )4
21 2dx
x −∫
( ) ( )2 4
2 21 2and
2 2dx dx
x x− −∫ ∫
( )211
1 1 12 22
cc dx
x cx = − = − − →∞ − − −∫
( )2
21 2dx
x −∫ diverges and shows that (∗) diverges.
( )
44
211
1 32 22
dxxx
= − = − − −∫