CHAPTER 10 Limits of Trigonometric Functions
CHAPTER 10
Limits of Trigonometric Functions
Some limits involve trigonometric functions. This Chapter explains how to deal with them. Let's begin with the six trigonometric functions.
10.1 Limits of the Six Trigonometric Functions
We
start
with
the
simple
limit
lim
x!c
sin(
x).
Here x is a radian measure because
c
we are taking sin of it. And because
x
the radian measure x approaches c, we interpret c as a radian measure too.
sin(x) sin(c)
The picture on the right illustrates this.
The point x on the unit circle moves to-
ward the point c on the circle. As this
happens, sin(x) approaches the number
sin(c).
Thus
lim
x!c
sin(x)
= sin(c).
For
example,
lim
x!
? 4
sin(x)
=
sin
? 4
?
=
p2 2
.
With a slight adaption, the above
picture
also
shows
lim
x!c
cos(
x)
=
cos(
c).
And
applying
limit
law
5,
we
get
lim
x!c
tan(x)
=
lim
x!c
sin(x) cos(x)
=
lim
x!c
sin(x)
lim
x!c
cos(x)
=
sin(c) cos(c)
=
tan(c),
provided
that
cos(c)
6=
0,
that
is,
c
6=
? 2
+ k?,
where
k
is
an
integer.
In
this
way
that we get the following formulas.
lim
x!c
sin(x)
=
sin(
c)
lim
x!c
cos(x)
=
cos(
c)
lim
x!c
tan(x)
=
tan(c)
lim
x!c
sec(x)
=
sec(
c)
lim
x!c
cot(x)
=
cot(
c)
lim
x!c
csc(x)
=
csc(
c)
for all real numbers c
for all real numbers c
for
all
real
numbers
c
6=
? 2
+ k?
for
all
real
numbers
c
6=
? 2
+ k?
for all real numbers c 6= k?
for all real numbers c 6= k?
148
Limits of Trigonometric Functions
Example 10.1
Find
lim
x!?
cos(x) x2
.
Because the denominator does not approach zero, we can use limit law 5
with
the
rules
just
derived.
Then
lim
x!?
cos(x) x2
=
lim
x!?
cos(x)
lim
x!?
2
x
=
cos(?) ?2
=
?1 ?2
.
Example 10.2
Find
lim
x!?/4
8x
tan(x) ? 2? 4x ??
tan(x)
.
Here
lim 8
x!
? 4
xtthaen(dx4e)x?n?o2?m? tianna(txo)r=axlp!imp?4r2oatacnh4(exxs)??4z?exr?o?, ?s=o
we
lim
x!
? 4
try to factor and cancel:
2
tan(x)
=
2
tan
? 4
?
=
2.
10.2 The Squeeze Theorem and Two Important Limits
It is easy to imagine limits where factoring and canceling is impossible, or
for which the limit laws do not apply. For example, in lim sin(x) we can't
factor
an
x
from
the
top
to
cancel
the
x
on
the
bottom
x!0
(which
apxproaches
0).
Actually, this particular limit turns out to be significant in calculus. We
now discuss a theorem that handles limits such as this one. The idea is to
cleverly compare a complicated limit to two simpler limits.
Theorem 10.1 (The Squeeze Theorem)
Suppose
we
need
to
compute
lim
x!c
g(x).
Suppose
also
that
we
can
find
two
functions f (x) and h(x) for which f (x) g(x) h(x) for values of x near c,
and
for
which
lim
x!c
f (x) =
L=
lim
x!c
h(x).
Then
lim
x!c
g(x)
=
L.
y
y = h(x)
y = g(x) L
y = f (x) x
c
The above picture illustrates the squeeze theorem. The graph of g(x) is
squeezed between the graphs of f (x) and h(x), both of which approach L as
x approaches c. The squeeze theorem states the obvious fact that in this
situation we can conclude that g(x) approaches L too.
Applying
the
squeeze
theorem
to
find
lim
x!c
g(x)
requires
some
ingenuity.
We have to find two other functions f (x) and h(x) for which f (x)g(x)h(x)
and
both
lim
x!c
f
(x)
and
lim
x!c
h(x)
are
easy
to
compute
and
are
both
equal
to
the
same
number
L.
At
that
point
the
squeeze
theorem
says
lim
x!c
g(x) =
L.
The Squeeze Theorem and Two Important Limits
149
We
will
next
use
the
squeeze
theorem
to
find
lim
x!0
sin(x) ,
x
which
will
be
needed in Chapter 21. But first let's think about what we'd expect it to be.
The unit circle on the right shows a radian mea-
sure x, close to 0. the vertical side of the tri-
angle is the corresponding value sin(x). Both
sin(x) and x are small, but the curved arc x
is so small that it looks almost like a vertical
line. The smaller x, the more "vertical" it looks,
and in fact it becomes almost indistinguishable
from the vertical side sin(x). For very small x the ratio sin(x) appears to be quite close to 1.
We
might
x
guess
lim
sin(x)
=
1.
x!0 x
sin(x) x
In fact, this turns out to be exactly the case. Proving it with the squeeze theorem requires a formula from geometry. Recall that a sector of a circle is a "pie slice" of the circle, as illustrated below, shaded.
Formula: The area of a sector of a circle of angle x and radius r is
A
=
1 2
2
r
x.
Here is why the formula works: The area of a
circle of radius r is ?r2. The sector takes up
only a fraction of this circle, that fraction being
x radians out unit circle, or
of 2? radians
x 2?
.
Thus
A
=
around
?r2
?
x 2?
=th12ere2 nx.tire
x r
A
1
We are ready to carry out our plan of proving
that will
lim sin(x) = 1 x!0 x
concoct two
via the squeeze theorem. We functions f (x) and h(x) with
f (x) sin(x) h(x) x
and
lim
x!0
f
(x)
=
1
=
lim
x!0
h(x).
The functions f and h will come from the dia-
gram on the right showing a sector OCP on the
unit circle, and another sector O AB of radius
cos(x) inside it. From this we get the following:
8 < sin(x) :
P
B x
O
| {z } cos(x)
A
C
150
Limits of Trigonometric Functions
!
!
!
Area of sector O AB
Area of triangle OCP
Area of sector OCP
.
Using the area formula for a sector (from the previous page) and the area formula for a triangle (from heart), this becomes
1 2
?
cos2(x)
?
x
1 2
?
1
?
sin(x)
1 2
?
12
?
x.
Actually, this only works if x is positive. If it were negative, then the above "areas" would be negative too. We correct this by taking the absolute value of the potentially negative terms x and sin(x).
1 2
?
cos2(x)
?
|x|
1 2
?
1
?
|
sin(x)|
1 2
?
12
?
|x|.
Now
multiply
all
parts
of
this
inequality
by
the
positive
number
2 |x|
to
get
cos2(x)
| sin(x)| |x|
1.
At this point the absolute values are unnecessary because if x is close to zero (as it is when x ! 0), then x and sin(x) are either both positive or both negative, so sin(x) is already positive. Updating the above, we get
x
cos2(x) sin(x) 1. x
Now
we've
squeezed
y=
sin(x) x
between
the
functions
y = cos2(x)
and
y = 1.
y
y = cos2(x)
0
y=1
y
=
sin(x) x
y = cos2(x)
Because
lim
x!0
cos2(x)
=
cos2(0)
=
1
=
lim 1,
x!0
the
squeeze
theorem
guarantees
lim sin(x) = 1. x!0 x
(10.1)
From
this
day
forward,
remember
the
fundamental
fact
lim
x!0
sin(x) x
=
1.
The Squeeze Theorem and Two Important Limits
151
y
=
Below cos2(x)
is a more complete picture of this situation, showing and y = 1. Notice that it's not the case that cos2(x)
y
= sin(x) sin(xx)
x
with 1 for
every value of x. But this does hold when x is near zero, and that is all we
needed to apply the squeeze theorem.
1
y = sin(x)
x
x ?5? ?4? ?3? ?2? ??
?
2?
3?
4?
5?
x
Students often assert incorrectly that sin(x) = 1. But that plainly wrong.
The
above
graph
shows
that
sin(x) x
never
x
equals
1.
In
fact,
sin(x) x
................
................
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