Use Limit to find the area and circumference of a circle
A = r2 , C = 2r
Yue Kwok Choy
1.
Prove that :
sin 
 1.
 0

lim
Proof:
>0.
Case 1 :
As in the diagram, AC is the tangent to the circle.
AC = r tan 
, where
r is the radius of the circle.
Area of OAB < area of sector OAB < area of OAC
1
1
1
 r 2 sin   r 2   r 2 tan 
2
2
2
 sin     tan 
1
Since
lim
0
lim
lim
0
1

sin 
 1, by Squeezing Principle, lim
 1  lim
1 .
0 sin 
0
cos 

sin 
sin   
 sin 
sin 
 lim
 lim
 lim
1
 0
   0   0 

Combine
2.
(on dividing by sin )
 < 0 , put  = –  , then  > 0 .
Case 2 :
0 

1

sin  cos 
Case 1 and Case2 , we have
lim
0
sin 
1 .

1  cos 
cos   1
 lim
0


0


1  cos 
2 sin 2  / 2
  sin  / 2 

 sin  / 2 
lim
 lim
 lim 
 lim lim 

  0 1  0 .
0


0


0


0


0


2  /2 
2
 /2 
2
Proof:
3.
Area of an n-sided regular polygon inscribed in a circle with radius r
Proof:
4.
2 
 .
n 
Exercise
2
sin
2 
1 2
n  r 2  1  r 2
Area of a circle = lim n  r sin   r 2 lim
n 
n 
2
n 
2
n
Perimeter of an n-sided regular polygon inscribed in a circle with radius r
Proof:
6.
1
2
= n r 2 sin
Area of a circle with radius r = r2 .
Proof:
5.
2


Exercise
Circumference of a circle with radius r = 2r .
Proof:

n
= n 2r sin  .

sin



n  2r  1  2r
Circumference of a circle = lim n 2r sin   2r lim
n 
n 

n


n