Theorems

Front 1

1.(an+bn)(n in N) converges to lim(an)+lim(bn)

Back 1

-let E>0.
Since lim(an)= A, we may chose an N1 in N such that for al n in N with n>=N1, we have |A-an|<E/2
Since lim(bn)=B, we may chose an N2 in N such that for n in N with n>=N2, we have |bn-B|<E/2
Set N=max(N1,N2) for all n in N with n>=N
We have |A+B-(an+bn)|=|(A-an)+(B-bn)|
By Triange inequality, <=|A-an|+|B-bn|<E/2+E/2=E

Front 2

2.(anbn)(n in N) converges to lim(an)lim(bn)

Back 2

Proof 2:
Choose E>0
Since (an) is a convergent sequence is it bounded, thus there is an M such that |an|<M for all n.
Now E’= E/(|B|+M)>0
Hence there is a positive Ni in N such that n>=N1 implies |an-A|<E’ and a positive integer N2 such that n>=N2 implies that |bn-B|<E’
Let N=max(N1,N2). For n>=N, we have:
|anbn-AB|=|(anbn-anB)+(anB-AB)|
By Tirange Inequality,
<=|an||bn-B|+|B||an-A|<=M|bn-B|+|B||an-A|<mE’+|B|E’=E
Thus anbn converges to AB

Front 3

If bn!=0, for an n in N and limbn)!=0, then (an/bn)(n in N), converges to (lim(an)/lim(bn))

Back 3

Because of part 2, it suffices to show that 1/bn converges to 1/B, where B=lim(bn)
Let E>0 be given.
By lemma 11, there exists an N1 in N and c>0, such that for all n in N with n>=N1,
|bn|>c.
Since bn converges to B, there exits N2in N such that for all n in N with n>=N2, we have |bn-B|<|B|Ec
Take N=max(N1,N2) and let n in N with n >=N .
Then |1/bn-1/B|=|B-bn|/|B||bn|
Because bn>c, then |B-bn|/|B||bn|<|B-bn|/c|B|
Because n>=N2, |B-bn|/c|B|<Ec|B|/c|B|=E

Front 4

Let an<=bn for all n in N
Suppose an converges to A and bn converges to B, then A<=B

Back 4

Proof:
By contraditicion
Assume WLOG A>B
Take E=(A-B)/2
Then E>0
By hypothesis, there exists N in N such that for all n in N with n>=N, we have |an-A|<E, and |bn-B|<E
Then an>A-E=B+E>bn
an>A-E=A+B/2+B+E>bn
Then an>bn, contradiction to hypthesis

Front 5

Let (an)(n in N) be a sequence that converges to 0 and let bn be a bouded sequence, then anbn converges to 0.

Back 5

|bn|<=M>0
E>0 and E’=E/M
Convergence of A
Then |anbn-0|=|anbn|=|an||bn|<=|an|M<E’M=E

Front 6

A sequence converges (1) iff each of its subsequences converges (2). In fact, if every subsequence converges, then they all converge to the same limit.
1=>2

Back 6

Proof
-(Assume 1 holds)Suppose (an)(n in N) converges to A.
-(Introduce (ank) and A) Let (ank) be a subsequence of (an). The limit of (an) is A.
-To show (ank) converges to A, let there be an E>0.
- (Def of convergence on (an)) Since lim(an) is A, then we can find there is an N in N such that for all n in N with n>=N we have |an-A|<E.
-(introduce k) Let k in N with k>=N, then n>=k>=N. Then we have |(ank)-A|<E. Therefore, (ank) converges to A)

Front 7

Let (an)(n in N) be a monotone sequence, then TFAE (an) converges and (an) is a bounded sequence
Decreasing

Back 7

Proof 12:
-it is true since all convergent sequences have been proven to be bounded (1.2)
Proof 2
-Suppose 2 holds
-Define s:=(an| n in N) to be bounded.
-( since s is decreasing, get an inf) Let m=inf of s.
- Then m+E is not a lower bound of s.
- (introduce E)Let E>0 be given, then m+E>m and hence m+E is not a lower bound of s.
-(introduce aN)This means that there is an N in n such that m+E>aN>=m
-By hypothesis, (an) is decreasing for all n in N with n>=N, we have
M+E>aN>=an>=m
-(an) is in( [m,m+E) C= (m-E,m+E))
Hence, |an -m|<E for all n in N with n>=N.

Front 8

Let (an)(n in N) be a monotone sequence, then THAE (an) converges and (an) is a bounded sequence
increasing

Back 8

Proof 21:
-Suppose 2 holds
-Define s:=(an| n in N) to be bounded
-Let M=sup of s.
-Let there be E>0, then M-E is not an upper bound and M-E<M.
-This means that there is N in N such that M-E<aN
-By hypotheses, (an) is an increasing sequence, and then we have
M-E<aN<=an<=M<M+E
-Therefore, |an-M|<E for all n in N with n>=N.

Front 9

Lef f:D R be a function, D in F, xo in R, xo an accu point of D.
Then TFAE f has a limit at xo and whenever (sn)(n in N) as a sequence converging to xo with (xn) in D\{xo}, for all n in N , she sequence f(xn)(n in N) also converges.
Supplement of 16: if and an 2 hold, then any sequence (xn) in D\{xo}, converges to the limf(x)

Back 9

Proof 12 and supp:
-Let (an)(n in N)be a sequence in D\{xo} which converges to xo.
-To prove f(xn)(n in N) converges to L, let E>0 be given. Since L=lim f(x), then there is a d such that for |x-xo|<d, then |f(xo)-L|<E.
-Sice d>0, then there is a N in N such that for all n>= N, |xn-xo|<d, then we deduce |f(xn)-L|<E by choice of d.

Proof 21:
-Let 2 hold True
-Step 1:
-Whenever (xn)(n in N) and (yn)(n in N) are sequences in D\{xo} and converge to xo, then the sequences f(xn) and f(yn) converge to the same limit xo.
-Define := (zn) (n in N) as follows:
-z2k=xk; z2k-1=yk (k in N)
-Thus zn is a sequence in D\{xo}, which converge to xo.
-Hence by 2, f(zn) converges to xo.
Step 2:
-Let L be all the coinciding sequences of f(xn)(n in N) where (xn) (n in N) is in D\{xo}
-Suppose L is not the limit of f at xo
-This means we can have an E>0 such that for all d>0, there exists an x in D\{xo}, but |F(x)-L|>=E.
-hence for each n in N, there exits an n in N with xn in D\{xo} with the property |xn-xo|<1/n and |f(xn)-L|>=E.
-Then xn is a sequence in D\{xo}. Let E>0 be given, then there is N in N such that for all n>=N, 1/n<E, then |xn-xo|<E and thus xn converges to xo.
-Because |f(xn)-L|>=E , the sequence does not converge to L.
-Contradiction to hypothesis

Front 10

:D--.R, D in R, xo in R, xo an accu point of D.
-Suppose f has a limit at xo, Then there exists a nbhd Z of xo and M in R such that |f(x)|<M for al x in Q∩D

Back 10

-Let L be the limit of F(x) and take E=1
-Then there is d>0, such that for al x in D\{xo}, |x-xo|<d, then |f(x)-L|<1
-Define Q:=(xo-d, xo+d), clearly Q is a nbhd of xo.
-case 1:
-When xo in D
-Then for M=max(|L-1|, |L+1|, |f(x)|)
-|f(x)|<M
-case2:
When xo not in D
-Then for M=max(|L-1|,|L+1|)
-f(x) is in (lL-1,L+1), therefore f(x)<M

Front 11

. f +g has limit A+B at xo

Back 11

Let (xn)(n in N) be a sequence with x in D\{xo}, which converges to xo.
-By theorem 16, we have f(xn) converges to A and g(xn) converges to B
-By theorem 10 ne have f(xn)+g(xn) converges to A +B
then by 16 again we have f+g=lim A+B

Front 12

f *g has limit AB at xo

Back 12

Let E>0
By thm 17, there is a nbhd Q of xo and M in R such that |f(x)|<M for x in QUnionD.
W.l.o.g., M>0.
Choose d1>0, such that there is (x-d1,x-d1) Cotained in Q.
By hypthesis there is a d2>0, such that for all x in D\{xo}Union(x-@2,x+@2), we have |g(x)-B|<E/2M+1
Since lim f(x)=A, then there exists @3, such that for al x in d\{xo}Union(x-d3,x+d30 we have |f(x)-A|<E/2|B|+1
Now take d=min(d1,d2,d3)
Then d>0 for all x in D\{xo} with |x-xo|<d we have:
|f(x)g(x)-AB|=|f(x)g(x) –f(x)B +f(x)B –AB|
By triangle inequality
<=|f(x)||g(x)-B|+|B||f(x)-A|,E/2+E/2=E

Front 13

if f(x)>0, for all x in D, then sqrt(f(x)) has limit sqrt(A) at xo

Back 13

If f(x)>0, for all x in D, and A>-0, then sqrt(f(x)) has limit sqrt(A) at xo.
A>=0 follows from 4 in thrm 18
To show limit(sqrt(f(x))) exists, and equals sqrt(A), let (xn)(n in N) be a sequence in D\{xo} converging to xo.
By hypothesis, lim(f(x)) = A and hence by thm 16, we can conclude lime(sqrt(f(x))) = sqrt(A)

Front 14

A sequence converges (1) iff each of its subsequences converges (2). In fact, if every subsequence converges, then they all converge to the same limit.


2=>1

Back 14

-Assume 2 holds true
-Since (an) is a subsequence of itself, then (an) converges by 2.