Post on 13-Jul-2015
By Ron English
EASY PROOF BUT
Only after you learn Derivatives and l’Hopital’s Rule
l'Hôpital's Rule
PROOF using
MATHEMATICS and
LIMITS!!!
You will need to know …⋆Squeeze theorem⋆Also known as the pinching theorem
⋆Area of arcs ⋆Area of an Arc of radius 1
⋆Area of triangles
⋆Trig functions
Squeeze Theorem
A B C≤ ≤if A C: =
∴ = =B A B C,
Also
TRIG of TRIANGLE
Adjacent
Opposite
Hypotenuse
x
MORE trig of a triangle
Adjacent
Opposite
Hypotenuse
x
AREA of a TRIANGLE RIGHT ANGLE TRIANGLE
HEIGHT
BASE
AREA =1/2 BASE * HEIGHT
AREA of an ARC
θ
RadiusArea = ½ x radius x θ
Start the PROOF
An Arc Radius 1
X
1
1
Draw lines from two intercepts
Add labels
XO A
B
C
D
Radius of Circle = 1 = OC = OB
THREE AREAS
XO A
B
C
D
Radius of Circle = 1 = OC = OB
SMALLEST AREA
XO A
B
C
D
Area = ½ base * height = ½ OA * AB
SMALLEST AreaArea = ½ base * height = ½ OA * AB
OA
BSin x = Opp / Hyp
= AB / OB
but: OB = 1
(radius of circle)
∴ Sin x = AB / 1
OR AB = Sin x
X
SMALLEST AreaArea = ½ base * height = ½ OA * AB
OA
BCos x = Adj / Hyp
= OA / OB
but: OB = 1
(radius of circle)
∴ Cos x = OA / 1
∴ OA = Cos x
X
SMALLEST AreaArea = ½ base * height = ½ AB * OA
OA
BAB = Sin x
OA = Cos x
X
Area = ½ Sin x * Cos x
½ Sinx*Cosx
Start the SQUEEZE Equation
MIDDLE Area
XO
B
CArea = ½ Radius * x
MIDDLE Area
XO
B
C
Area = ½ Radius * xRadius = 1
Area = ½ x
½ Sinx*Cosx <=½ x
Middle of SQUEZE Equation
LARGEST Area
XO
C
D
Area = ½ base * height = ½ OC * CD
LARGEST Area
XO C
DArea = ½ base * height = ½ OC * CD
OC = 1 (Radius of Circle)
LARGEST Area
XO
C
DArea = ½ base * height = ½ OC * CD
TAN x = DC / OC
since OC = 1
∴ TAN x = DC
but TAN x =Sin x / Cos x
LARGEST Area
XO
C
DArea = ½ base * height = ½ OC * CD
DC = TAN x
DC = Sin x / Cos x
Area = ½ Sin x / Cos x
Complete SQUEEZE Equation
Some Arithmetic
Some more Arithmetic
Even MORE Arithmetic
a LIMIT
Take LIMITS a X approaches Zero
QED
Latin for
quod erat demonstrandum
(which was to be demonstrated“)