Limits and Continuity for the Trigonometric Functions
sin(x)
1
x
cos(x)
x
cos(x)
sin(x)
Thus all trigonometric functions are continuous in their natural
domain.
One important tool for finding limits is the “squeezing theorem”
Theorem. Let f, g, and h be functions satisfying
g(x) ≤ f(x) ≤ h(x)
for all x in an open interval containing c, except possibly at c.
If:
then
lim g ( x) = lim h( x) = L
x →c
x→ c
lim f (x) = L
x →c
We will not prove this result, but it is clearly illustrated by the
following diagram.
3
h(x)
2
1
f(x)
-3
-2
1
-1
2
(0,0)
-1
g(x)
-2
-3
x=c
3
Example. Compare sin(1/x) and xsin(1/x) at 0.
y = sin(1/x)
As x tends to 0, 1/x tends to infinity. Thus, as x tends toward 0,
there are values of x for which 1/x is 2π, 4π, 6π, 8π, etc. At all of
these points, the sine is 0. There are also, as x tends toward 0,
values of x for which 1/x is π/2, (π/2) + 2π, (π/2) + 4π, (π/2) +
6π, etc., and at these points the sine is 1. We conclude that the
function sin(1/x) has no limit as x tends to 0.
The function xsin(1/x) has the following graph.
This function is very difficult to analyze directly, but we can use
the squeezing theorem. Since −1 ≤ sin( x) ≤ 1
it follows that − x ≤ xsin( x) ≤ x.
Since lim x = lim − x =0 we have lim x sin(1/ x) = 0
x →0 x →0
x →0
Two Important Trigonometric Limits
r
x
r
Area of green sector is 1 r 2 x
2
(1, tan(x))
(cos(x), sin(x))
x
sin(x)
tan(x)
1
1
1
sin( x)
Area =
2
1
x
Area =
2
Area = tan( x)
2
Thus sin(x) < x < tan (x), and so if we divide by sin(x), we have
x
1
1<
<
sin(x) cos( x)
or,after taking reciprocals
cos( x) < sin(x) < 1 (*)
x
for every x between 0 and π/2. This also holds between – π/2
and 0, since sin(–x) = sin(x) and cos(–x) = cos(x).
Thus (*) holds in the interval (– π/2 , π/2) around 0.
Since lim cos(x) = cos(0) =1 the squeezing theorem shows that
x →0
lim sin( x) =1
x →0 x
The second limit is found from this by using the identity
sin 2 (x )=1−cos 2 (x)
2 (x)
1
−
cos
1
−
cos(
x
)
1
−
cos(
x
)
1
+
cos(
x
)
= lim
lim
= lim
g
x
x
1+ cos(x)
x→0 x(1+ cos( x) )
x →0
x →0
2 ( x)
sin(x)g sin( x) = 0
sin
=
lim
= lim
x→0 x(1+ cos( x) ) x→0 x 1+ cos( x)
Exercise. Find any points of discontinuity for the function
f ( x) =
4
1− 2cos(x)
Solution. The function is continuous except where the
denominator is 0. This happens when 1 − 2cos(x) = 0, or
cos(x) = ½.
Thus x = π/3, –π/3 or these angles plus any multiple of 2π.
This is shown in the following picture.
cos(x)
Example. Compute
2
lim sin  
x→∞  x 
Solution. Since the sine is a continuous function, we know by a
previous theorem that
2
2

lim sin   = sin  lim  = sin(0) = 0
x→∞  x 
 x →∞ x 
Example. Compute
lim sin(h)
h→0 5h
Solution. Since constants can come out of the limit, we have
lim sin(h) = 1 lim sin(h) = 1
5
h→0 5h 5h→0 h
Example.Compute
lim sin(5s)
s→0 s
Solution. We make a substitution, u = 5s. Then
lim sin(5s) = lim 5sin(5s ) = 5 lim sin(5s) =5lim sin(u) = 5
s →0 s
s →0 5s
s →0 5s u →0 u
Example. Compute
 θ2
lim 
θ →0 1− cosθ




Solution. This expression must be changed to be understood.
We will multiply the top and bottom by 1 + cos(θ)
 θ2
lim 
θ →0 1− cosθ

 θ2
 = lim 
 θ →0 1−cosθ


 1+ cosθ 

 1+ cosθ 

 θ 2(1+cosθ ) 
 θ2
= lim 
 = lim (1+ cosθ ) 2
2
 sin θ
θ →0 1−cos θ  θ →0

 θ2
= lim (1+ cosθ ) lim  2
θ →0
θ → 0 sin θ

=2






Example. Compute
 h 
lim 

h →0 tan(h) 
Solution. Here we begin with a trigonometric identity.
h
 h 


 h cos(h) 
lim 
=
lim
=
lim





h →0 tan(h)  h →0 sin(h) / cos(h)  h →0 sin(h) 
h 

=  lim cos(h)  lim
=1

 h →0
 h →0 sin(h) 
Example. Compute
 t +3sin t 
lim 

t
t →0

Solution. Here we begin by breaking up a sum.
t 
3sin t 
 t + 3sin t  
lim 
=  lim  + lim


t
t
t
t →0
  t →0   t →0

sin t 

=1+ 3 lim
=4

 t →0 t 