Chapter 2 - LIMITS/Chapter 3 - Continuity of Functions

Question 1

Number Sequence(c2)

Answer 1

n0 as default( starter)
-> (an) with n>=n0 as (an) are numbers in the sequence
Therefore we have a number of sequences as an0,an0+1,an0+2,…

Question 2

Bounded/Unbounded Sequences(c2)

Answer 2

front end/back end (there exists a K that 0<K<infinity with an=<K- Bounded
Doesn’t have a front or a back end

Question 3

Convergent/Divergent(c2)

Answer 3

Finite limit/Unlimited

Question 4

Limit Laws(c2) (Cauchy and Squeeze)

Answer 4

• Cauchy criterion
  With every e>0, exist N=N(e) :n,m>N(e) =>[an-am]<e
• Squeeze theorem
  an<=bn<=cn with n>=m>=1, and liman=limcn=L as n head to infinity
  => limbn as n head to infinity also =L

Question 5

Hyperbolic Functions(c2)

Answer 5

sinhx=(e^x-e^(-x))/2 (odd)
coshx=(e^x+e^(-x))/2 (even)
with - infinity<x<infinity

Question 6

Infinites and Infinitesimals and Comparison of Infinitesimals and Infinities(c2)

Answer 6

• a(x) is called a infinitesimal as x approaches x0 if lima(x)=0 with x->x0
• b(x) is called a infinity as x approchaches x0
  if limb(x)=infinity with x->x0
  Comparison of infinitesimal
• Let a1(x) and a2(x) be two infinitesimals as x approaches x0.Let lim (a1(x)/a2(x))=k as x->x0
  + if k=0 then a1(x) is of higher order of a2(x) denote
  a1(x)=o(a2(x)) as x->x0
  + if 0<[k]<infinity then a1(x) is of the same order of a2(x), denote
  a1(x)=O(a2(x)) as x->x0
  + if k=1 then a1(x) and a2(x) are called to be equivalent, denote
  a1(x) ~ a2(x) as x->x0
• Let b1(x) and b2(x) be two infinities as x->x0. Let lim (b1(x)/b2(x))=k as x->x0
  +if k=0 then a1(x) is of higher order of a2(x) denote
  a1(x)=O(a2(x)) as x->x0
  +if 0<[k]<infinity then a1(x) is of the same order of a2(x), denote
  a1(x)=o(a2(x)) as x->x0
  + same as infinitesimal when they are equivalent.

Question 7

Particular Limits(c2)
Imported pairs of equivalent infinitesimal
as x->0

Answer 7

set a belongs to the complex of {sinx, tanx, e^x-1, ln(x+1), arcsinx, arctanx
then lim(a)~x as x tend to 0
Special cases: 1-cosx ~ x^2/2
e^u~u
a^x -1~xlna
(1+x)^a-1~ax

Question 8

Continuity

Answer 8

Definition: Suppose there is an open interval (a,b) which contains x0 then f(x) is continuous at x0 if lim f(x)=f(x0) as x->x0
Theorem:
!If the function f(x) is continuous on [a,b] then
For every e>0, exists G=G(e) : for every x’,x’’ belongs to [a,b], [x’-x’’]<G => [f(x’)-f(x’’)]<e
We say then f(x) is uniformly continuous on [a,b]. (THIS TERM IS COMPLETELY OPTIONAL TO REMEMBER AS IT IS A ADVANCED TOOL TO SOLVE PROBLEMS)
Theorem 1:
If both f(x) and g(x) are continuous at x0 then the following functions are also continuous at x0:
1.f + g
2.f - g
3. fg
4.cf for any constant c
5.f/g if g(x0) different than 0
Theorem 2:
If g(x) is continuous at x0 and f(x0) is continuous at g(x0) then the composite function f(g(x0)) is continuous at x0.
Theorem 3:
If f(x) is continuous at b and lim f(x)=b as x->x0 , then lim g(f(x))=g(lim f(x))=g(b) as x->x0
Theorem 4:
The essential functions are continuous in their domains.
Theorem 5:( The intermediate Value Theorem)
Suppose that f(x) is continuous on a CLOSE interval [a,b] then f(x) attains the smallest value m and largest M. And for every number K (m<=K<=M), there exists a number c,a<=c<=b such that f(c)=K

Question 9

Discontinuity

Answer 9

Suppose there is a x0 where the function f(x) is discontinued then we call x0 a discontinuity point. There are two types of discontinuity:
-Discontinuity type one is when lim f(x) as x tend to x0- is DIFFERENT from lim f(x) as x tend to x0+.
-Discontinuity type two is when lim f(x) as x tend to x0- is EQUIVALENT to limf(x) as x tend to x0+.
NOTE:Both of these lim has to EXIST and are FINITE.
-The removal of discontinuity is when at x0(the point of discontinuity) exists a limf(x)=L (a finite number) as x tend to x0