BC 1
Intro to Special Limits Involving Trig Functions
Consider the following limits: lim
 0
sin  

Name:
and lim
 0
1  cos  

.
What do you think the value of each of these limits should be? A quick check of the graphs of
1  cos  x 
sin  x 
and y 
suggest possible answers.
y
x
x
y
sin x
x
y
Based upon these graphs, it appears lim
 0
sin  

 1 and lim
1  cos  
 0

1  cos x
x
 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 interval that contains x = a.
If lim (x) = L and lim h  x   L , then it must be the case that lim g  x   L
x a
xa
Proof of lim
 0
sin  

xa
 1:
We consider the case where  > 0. Since the function    
true?), the case of  < 0 is taken care of as well.
sin  

is an even function (why is this
Intro to Special Limits Involving Trig Functions.1
F11
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:
T
P
(Area of OAP) ≤ (Area of sector OAP) ≤ (Area of OAT)

Q
O
A(1,0)
We will use these inequalities and the Squeeze Theorem to prove
sin  
that lim
 1.

 0
(Area of OAP) < (Area of sector OAP) < (Area of OAT)
1
 OA  PQ
2
≤
1
1  sin  
2
≤
1
≤
1
≥
1
≥
Thus, lim
 0
sin  


2
  1
2
≤

2
≤
1
1  tan  
2
≤
1
cos  
≥
cos  

sin  
sin  

lim
sin  

 0
 0

PQ PQ

and
OP
1
AT AT
tan   

.
OA
1
2

Multiply by
sin 
Since sin   
Take reciprocals, which will
reverse all the inequalities.
Take limits.
 1 by the Squeeze Theorem.
 0
1  cos  
≥ lim cos    1
 0
This limit can be used to prove that lim
lim
1
 OA  AT
2
1  cos  

 0.
1  cos   1  cos  
1  cos 2  
sin 2  

 lim
 lim
 0

1  cos    0  1  cos     0  1  cos   
 lim
 lim
 0
sin  


sin  
0
 1  0
1  cos  
2
Intro to Special Limits Involving Trig Functions.2
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