BC 1
Limits 5 (intro)
Name:
Intro to Special Limits Involving Trig Functions
Consider the following limits: lim

0

and lim

0

.
What do you think the value of each of these limits should be? A quick check of the graphs of y
 sin
 
and y

1 cos
 
suggest possible answers. x x y
 sin x x y

1 cos x x
Based upon these graphs, it appears lim

0 sin


1 and lim

0


0 . Now we know graphs only provide convincing evidence and do not “prove” anything, so how can we prove the values for either of these limits are correct. One possibility is to make use of the Squeeze Theorem.
Squeeze Theorem
Suppose that

( x
) ≤ g ( x
) ≤ h ( x ) for all x in an open interval that contains x = a .
If lim x
 a

( x ) = L and lim x
 a
  
L , then it must be the case that lim x
 a
  
L
Proof of lim

0 sin


1 :
We consider the case where

> 0. Since the function sin
 

is an even function (why is this true?), the case of

< 0 is taken care of as well.
IMSA BC 1 Lim 5i.1 S12

O

P
Q
T
A(1,0)
In the figure at the left, a quarter circle with radius 1 has been drawn. Point P is a point on this circle. Considering the areas of

OAP,

OAT, and sector OAP, we can form the following inequalities:
(Area of
 OAP) ≤ (Area of sector OAP) ≤ (Area of 
OAT)
We will use these inequalities and the Squeeze Theorem to prove that lim

0 sin

 

1 .
(Area of

OAP) < (Area of sector OAP) < (Area of

OAT)
1
2
  ≤
2


1
2
   
≤

2
  2 ≤
1
2
 
≤
1
2
 
1
≤ sin

 
≤ cos
1
 
1
≥ sin
 

≥ cos
 
1
≥ lim

0 sin

≥ lim cos

0
 
1
Thus, lim

0 sin


1 by the Squeeze Theorem.
Since sin tan
PQ

PQ
OP 1
and
AT

AT
OA 1
.
Multiply by
2 sin


Take reciprocals, which will reverse all the inequalities.
Take limits.
This limit can be used to prove that lim

0 lim

0



 lim

0
1 cos

 

1 cos
 
  lim

0 sin

 sin
 
  1

2 lim

0
0

0 .

2
 
  
 lim

0
 sin
2
 
  
IMSA BC 1 Lim 5i.2 S12