4. Sequences of real numbers. Monotonic Sequences.
Answer:
We denote a sequence by {xn}oon=1. The real number xn is the n-th term of the sequence.
Definition. A sequence {xn} converges to a limit a if for every ε>0 there is an integer nε such that |xn-a|<ε, if n>nε
In this
case we say that {xn}
is convergent and we write
n=a.
A sequence that does not converge diverges or is divergent.
Theorem. The limit of a convergent sequence is unique.
Proof:
Suppose that
n=a
and
n=b.
We must show that a=b. Let ε>0.
From definition of limits, there are integers N1
and N2
such that |xn-a|<ε
if n>N1
(because
n=a);
|xn-b|<ε
if n>N2
(because
n=b)
This inequelities both hold, if n>N=max(N1, N2) which implies that |a-b|=|a-xn+xn-b|≤|a-xn|+|xn-b|=ε+ε=2ε
Since, this inequality holds for every ε>0 and |a-b|=0; that is a=b
We say that n= , if for any real number axn>a for large n
Definition. A sequence {xn} is bounded above if there is a real number b such that xn≤b for all n , bounded below, if there is a real number a such that xn≥a for all n , or bounded, if there is a real number r such that |xn|≤r (-r≤xn≤r) for all n
Monotonic Sequences.
Definition: A sequence {xn} is increasing, if xn<xn+1 ; non-decreasing, if xn≤xn+1 ; decreasing, if xn+1<xn; non-increasing, if xn+1≤xn.
Theorem. If a sequence {xn} is monotonic and bounded then it is convergent. Cauch’s Convergence Criterion. A sequence {xn} converges inӀRn if and only if for every ε>0 there is nεͼN, for every n>nε ,for every pͼN => |xn+p-xn|<ε.
Theorem.
Let
n=S
and
n=t
where S and t are finite. Then:
n=cS
, c is constant.
=S+t,
(the limit of is the sum of the limits)
n=S-t,
=St,
=
(if tn
is non-zero for all n
and t
0),(the
limit of quotient is the quotient of the limits, provided t
0)
Definition. The numbers Ṡ and S are called the limit superior and limit inferior respectively of {Sn} and denoted by Ṡ= n and S= n
Theorem.If
{an}
is a sequence of real number, then
n=b
if and only if |
n=
n=b
5.Functions and Limits. Properties of the limits.
Answer:
A rule that assings to each member of a nonempty set D a unique member of a set Y is a function from D to Y, and we write y= (x). The set D is the domain of , denoted by Df. The members of Y are the possible values.
Definition.
We say that
(x)
approaches the limit L as x approaches x0
, and write:
(x)=L
, (1)
iff is defined on some deleted neighborhood of x0 and for every ε>0 there is δ>0 such that |f(x)-L|<ε (2)
if 0<|x-x0|<δ (3)
Theorem.If (x) exists, then it is unique.
Proof:
Suppose that
L1
and
(x)=L2
(4)
Let ε>0. From Definition there are positive members δ1 and δ2 such that:
|f(x)-Li|<ε (5) if 0<|x-x0|<δi , i=1,2,3,….
If (7) δ=min(δ1, δ2), then |L1-L2|=|L1-f(x)+f(x)-L2|≤|L1-f(x)|+|f(x)-L2|<2ε
If 0<|x-x0|<δ
We have now established an inequality that does not depend on x; that is |L1-L2|<2ε
Since this is holds for any positive ε, L1=L2.
Theorem.
If
(x)=L1
and
(x)=L2
(9)
5.2 Then
(x)=L1+L2
(10)
(x)=L1-L2
(11)
(x)=L1L2
(12)
and if L2
0
(13)
(x)=
(14).
Proof: From (9) and Definition, if ε>0 there is δ>0 such that |f(x)-L1|<ε (15), if 0<|x-x0|<δ1 and a δ2>0 such that |g(x)-L2|<ε (16) if Suppose that 0<|x-x0|<δ=min(δ1,δ2) (17)
|(f
g)(x)-(L1
L2)|=|(f(x)-L1)
(g(x)-L2)|
|f(x)-L1|+|g(x)-L2|<2ε,
which proves (10) and (11). To prove (12), we assume (17) and
write
|(fg)(x)-L1L2|=|f(x)g(x)-L1L2|=|f(x)g(x)-f(x)L2+L2f(x)-L1L2|=|f(x)(g(x)-L2)+L2(f(x)-L1)|
|f(x)(g(x)-L2)|+|L2(f(x)-L1)|=|f(x)||g(x)-L2|+|L2||f(x)-L1|<(|f(x)|+|L2|)ε
(|f(x)|-|L1|+|L1|+|L2|)
(|f(x)-L1|+|L1|+|L2|)
(ε+|L1|+|L2|)ε
(1+|L1|+|L2|)ε
if ε<1 and x satisfies (17). This proves (12). To prove
(14), we first observe that if L2
0
there is δ3>0
such that |g(x)-L2|<
,
so g(x)>
(18) if 0<|x-x0|<
.
To see this, let L1=L2
and ε=
.
Now suppose that 0<|x-x0|<min(δ1,δ2,δ3)So
that (15), (16) and (18) all hold. Then
|(
)(x)-
|=|
-
|=
|L2f(x)-L1g(x)|=
|L2(f(x)-L1)+L1(L2-g(x))|
(|L2||f(x)-L1|+|L1||L2-g(x)|)
(|L2|+|L1|).
This proves (14)