Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
2008
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
Ch2-4: Some Limit Theorems
Theorem (Uniqeness of a Limit) If lim
x→cf(x) =L and lim
x→cf(x) =M , then L=M.
Pf:
Assume that L6=M Since lim
x→cf(x) =L,∃δ1>0 s.t.
if 0<|x −c|< δ1, then
|f(x)−L|< ǫ≡ |L−M|
2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
And∃δ2>0 s.t.
if 0<|x −c|< δ2, then
|f(x)−M|< ǫ Let x1 satisfy 0<|x1−c|<min{δ1, δ2}, then|f(x1)−L|< |L−M|2 ,|f(x1)−M|< |L−M|2
⇒ |L−M| ≤ |L−f(x1)|+|f(x1)−M|
<|L−M| Contradiction. 2
WEN-CHINGLIEN Calculus (I)
The Sandwich Theorem
Theorem (Comparison)
If f(x)≤g(x),∀x ∈(a,b)and x 6=c , c ∈(a,b) then lim
x→cf(x)≤ lim
x→cg(x) Theorem
If f(x)≤g(x)≤h(x),∀x ∈(a,b)and x 6=c , c∈(a,b) and lim
x→cf(x) = lim
x→ch(x) =L, then lim
x→cg(x) =L
WEN-CHINGLIEN Calculus (I)
The Sandwich Theorem
Theorem (Comparison)
If f(x)≤g(x),∀x ∈(a,b)and x 6=c , c ∈(a,b) then lim
x→cf(x)≤ lim
x→cg(x) Theorem
If f(x)≤g(x)≤h(x),∀x ∈(a,b)and x 6=c , c∈(a,b) and lim
x→cf(x) = lim
x→ch(x) =L, then lim
x→cg(x) =L
WEN-CHINGLIEN Calculus (I)
The Sandwich Theorem
Theorem (Comparison)
If f(x)≤g(x),∀x ∈(a,b)and x 6=c , c ∈(a,b) then lim
x→cf(x)≤ lim
x→cg(x) Theorem
If f(x)≤g(x)≤h(x),∀x ∈(a,b)and x 6=c , c∈(a,b) and lim
x→cf(x) = lim
x→ch(x) =L, then lim
x→cg(x) =L
WEN-CHINGLIEN Calculus (I)
The Sandwich Theorem
Theorem (Comparison)
If f(x)≤g(x),∀x ∈(a,b)and x 6=c , c ∈(a,b) then lim
x→cf(x)≤ lim
x→cg(x) Theorem
If f(x)≤g(x)≤h(x),∀x ∈(a,b)and x 6=c , c∈(a,b) and lim
x→cf(x) = lim
x→ch(x) =L, then lim
x→cg(x) =L
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
Example1: lim
x→0x sin1 x Example2: lim
x→0
sin x Pf: x
0<x < π 2, BD=⌢ x, OA=cos x, AD =sin x, BC =tan x.
Area △OAD≤ Area∢OBD ≤Area △OBC
⇒ 12cos x sin x ≤ 121x ≤ 121 tan x
WEN-CHINGLIEN Calculus (I)
⇒ cos x ≤ x
sin x ≤ 1 cos x
⇒ 1
cos x ≥ sin x
x ≥cos x
⇒ lim
x→0
sin x
x =1. 2
WEN-CHINGLIEN Calculus (I)
⇒ cos x ≤ x
sin x ≤ 1 cos x
⇒ 1
cos x ≥ sin x
x ≥cos x
⇒ lim
x→0
sin x
x =1. 2
WEN-CHINGLIEN Calculus (I)
⇒ cos x ≤ x
sin x ≤ 1 cos x
⇒ 1
cos x ≥ sin x
x ≥cos x
⇒ lim
x→0
sin x
x =1. 2
WEN-CHINGLIEN Calculus (I)
Exercise:
1 lim
x→∞
ln x x =?
2 lim
x→0
sin(2x)
3x =?
3 lim
x→0
1−cos x
x =?
4 lim
x→0x2sin1 x =?
WEN-CHINGLIEN Calculus (I)
Exercise:
1 lim
x→∞
ln x x =?
2 lim
x→0
sin(2x)
3x =?
3 lim
x→0
1−cos x
x =?
4 lim
x→0x2sin1 x =?
WEN-CHINGLIEN Calculus (I)
Exercise:
1 lim
x→∞
ln x x =?
2 lim
x→0
sin(2x)
3x =?
3 lim
x→0
1−cos x
x =?
4 lim
x→0x2sin1 x =?
WEN-CHINGLIEN Calculus (I)
Exercise:
1 lim
x→∞
ln x x =?
2 lim
x→0
sin(2x)
3x =?
3 lim
x→0
1−cos x
x =?
4 lim
x→0x2sin1 x =?
WEN-CHINGLIEN Calculus (I)
Thank you.
WEN-CHINGLIEN Calculus (I)
