Continuity of Trig and Inverse
Functions
Objective: To use limits to define
continuity in trig/inverse functions
Theorem 1.6.1
• If c is any number in the natural domain of the stated
trigonometric function, then
lim sin x  sin c
lim cos x  cos c
lim tan x  tan c
lim cot x  cot c
x c
x c
lim sec x  sec c
lim csc x  csc c
x c
x c
x c
x c
Example 1
• Find the limit
 x2 1 

lim cos
x 1
 x 1 
Example 1
• Find the limit
 x2 1 

lim cos
x 1
 x 1 
• The limit of the cosine is the cosine of the limit.


 x2 1 
 x2 1 
  cos lim 
  cos lim ( x  1)  cos 2
lim cos
x 1
x 1
x 1
 x 1 
 x 1 
Theorem 1.6.2
• If f is a one-to-one function that is continuous at each
point of its domain, then f -1 is continuous at each
point of its domain, that is f -1 is continuous at each
point of the range of f.
Example 3
• Where is the function
continuous?
tan 1 x  ln x
f ( x) 
x2  4
Example 3
• Where is the function
continuous?
tan 1 x  ln x
f ( x) 
x2  4
• A fraction is continuous where the numerator and
denominator are continuous and the denominator is
not zero.
Example 3
• Where is the function
continuous?
tan 1 x  ln x
f ( x) 
x2  4
• A fraction is continuous where the numerator and
denominator are continuous and the denominator is
not zero.
• The numerator is continuous for x > 0 (why?) and the
denominator is continuous everywhere x  2
Example 3
• Where is the function
continuous?
tan 1 x  ln x
f ( x) 
x2  4
• A fraction is continuous where the numerator and
denominator are continuous and the denominator is
not zero.
• The numerator is continuous for x > 0 (why?) and the
denominator is continuous everywhere x  2
• The function f is continuous for x > 0, not 2.
The Squeezing Theorem
• Let f, g, and h be functions satisfying
g ( x )  f ( x )  h( x )
for all x in some open interval containing the
number c, with the possible exception that the
inequalities need not hold at c. If g and h have the
same limit as x approaches c, say
lim g ( x)  lim h( x)  L
x c
x c
then f also has this limit
f ( x)  L
as x approaches c, that is lim
x c
Theorem 1.6.5
sin x
( a ) lim
1
x 0
x
1  cos x
(b) lim
0
x 0
x
Example 4
• Find:
tan x
lim
x 0
x
Example 4
• Find:
tan x
lim
x 0
x
sin x 
1 
 sin x 1  
lim 

   lim
 lim
  1 1  1
x 0
x

0
x

0
x 
cos x 
 x cos x  
Example 4
• Find:
sin 2 x
lim
x 0
x
Example 4
• Find:
sin 2 x
lim
x 0
x
2 sin 2 x
sin 2 x
lim
 2 lim
 2 1  2
x 0
x 0
2x
2x
Example 4
• Find:
sin 3 x
lim
x  0 sin 5 x
Example 4
• Find:
sin 3 x
lim
x  0 sin 5 x
sin 3 x
lim x
x  0 sin 5 x
x
Example 4
• Find:
sin 3 x
lim
x  0 sin 5 x
sin 3x
sin 3x
3
3x
lim x  lim
x 0 sin 5 x
x 0
sin 5 x
5
x
5x
Example 4
• Find:
sin 3 x
lim
x  0 sin 5 x
sin 3x
sin 3x
3
3
x
3
x
lim
 lim
 lim
x 0 sin 5 x
x 0
sin 5 x 5 x0
5
x
5x
3 1 3
 
5 1 5
sin 3x
3x
sin 5 x
5x
Homework
• Section 1.6
• Pages125-126
• 1-35 odd