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 lxi!mc sin(x). Here x is a radian measure because
we are taking sin of it. And because the radian measure x approaces c, we interpret c as a radian measure too.
c x
x
c sin( )
sin( )
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,
x sin( )
approaches
the
number
sin(c). Thus lxi!mc For example,
x= sin( )
x
sin(c). ?
=
?
=
p 2
.
With a slight adaption, the above
xl!im? sin( ) sin
42
picture
also
4
shows
lxi!mc
x cos( )
=
cos(c).
And
applying
limit
law
5,
we
get
lxi!mc
tan(
x )
=
x sin( ) lxi!mc x cos( )
=
lxi!mc
x sin( )
lxi!mc
x cos( )
=
c sin( )
c cos( )
=
c tan( ),
provided that c 6= , that is, c 6= ? + k?, where k is an integer. In this way cos( ) 0
2
that we get the following formulas.
lxi!mc
sin(
x )
=
c sin( )
lxi!mc
x cos( )
=
c cos( )
lxi!mc
x tan( )
=
c tan( )
lxi!mc
x sec( )
=
c sec( )
lxi!mc
x cot( )
=
c cot( )
lxi!mc
csc(
x )
=
c csc( )
for all real numbers c
for all real numbers c
for all real numbers c 6= ? + k?
2
for all real numbers c 6= ? + k?
2
for all real numbers c 6= k?
for all real numbers c 6= k?
148
Limits of Trigonometric Functions
x
Example 10.1
Find
xli!m?
cos( x2
).
Because the denominator does not approach zero, we can use limit law 5
with
the
rules
just
derived.
Then
xli!m?
x cos( )
x2
=
xli!m?
x cos( )
xli!m? x2
=
? cos( )
?2
=
? 1
.
?2
Example 10.2
Find
xl!im?/4
x 8
x? tan( )
x?
x 2 tan( ) ?
.
4
Here
8 xl!im?
xtthaen(dxe)x?n?o2?m? tianna(txo)r=axlp!imp?r2oatacnh(exxs)??4z?exr?o?, ?s=o
we xl!im?
try to factor and cancel:
2
x tan( )
=
2
tan
?
?
=
2.
4
4
4
4
4
4
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
x sin( ) xli!m x
we
can't
factor
an
x
from
the
top
to
cancel
the
x
on
the
bottom
0
(which
approaches
).
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 lxi!mc 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
lxi!mc
fx= ()
L
= lxi!mc h(x).
Then
lxi!mc
gx ()
=
L.
y y=h x ()
L
y=g x ()
y= f x
c
( )x
The
above
picture
illustrates
the
squeeze
theorem.
The
graph
of
gx ()
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
gx ()
approaches
L
too.
Applying
the
squeeze
theorem
to
find
lxi!mc
gx ()
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
lxi!mc
f
x ()
and
lxi!mc
hx ()
are
easy
to
compute
and
are
both
equal
to
the
same
number
L.
At
that
point
the
squeeze
theorem
says
lxi!mc
gx ()
=
L.
The Squeeze Theorem and Two Important Limits
149
We
will
next
use
the
squeeze
theorem
to
find
xli!m
x sin( )
x
,
which
will
be
0
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
x sin( )
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 x . For very small x
the We
ratio might
sgiunx(exs)saxlpi!mpesainxr(ssxi)nt=(o
) be
1.
quite
close
to
1.
0
xx sin( )
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 r2x .
2
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
oxf
2? radians . Thus A =
?arr2ou? nxd
the entire = 1 r2x.
?
?
2
22
x
rA
1
We are ready to carry out our plan of proving
that
xli!m
x sin( )
x
= 1
via
the
squeeze
theorem.
We
will fx ()
cosn0incx(oxc)ttwh(ox)funacntdionsxli!mf (fx)(xa) n=d1
hx ()
with
= xli!m h(x).
The
functions
f
and
h
will
0
come
from
0
the
dia-
gram on the right showing a sector OCP on the
unit circle, and another sector O AB of radius
x cos( )
inside
it.
From
this
we
get
the
following:
8
<
x sin( )
:
P
B
x
O|
{z x
}
A
C
cos( )
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 x ?x 1? ? x 1? 2?x
cos ( )
1 sin( )
1.
2
2
2
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 x ? |x| 1 ? ? | x | 1 ? 2 ? |x|
cos ( )
1 sin( )
1.
2
2
2
Now multiply all parts of this inequality by the positive number |2x| to get
2x cos ( )
| x| sin( )
|x|
1.
At this point the absolute values are unnecessary because if x is close to
zero (as it is negative, so
swinxh(xe) nisxa!lre0a),dtyhepnosxitaivned.
x sin( )
are
Updating
either both positive the above, we get
or
both
2x cos ( )
x sin( )
x
1.
Now
we've
squeezed
y=
x sinx( )
between
the
functions
y= 2 x cos ( )
and
y = 1.
y
y= 2 x cos ( ) 0
y= 1
y
=
x sinx( )
y= 2 x cos ( )
Because
2x xli!m0 cos ( )
=
2
cos (0)
=
1
=
xli!m0 1,
the
squeeze
theorem
guarantees
x
sin(
xli!m
0
x
)= 1.
(10.1)
x
From
this
day
forward,
remember
the
fundamental
fact
xli!m
sin( x
) = 1.
0
The Squeeze Theorem and Two Important Limits
151
y
=
Below 2x
cos ( )
is a more complete picture of this situation, showing
and y = . Notice that it's not the case that 2 x
1
cos ( )
y
=sinxs(ixnx)(x)
with 1 for
value of x. But this
hold when x is near zero, and that is all we
every
does
needed to apply the squeeze theorem.
1
y
=
x sin( )
x
x
?5? ?4? ?3? ?2?
??
x
?
2?
3?
4?
5?
TheSatbuodveengtsraopfthenshaoswssertthaintcsoinrx(rxe) cntleyvtehraetqsuinxa(xl)s=1.1.InBfuatctt,hsaintx(px)la ................
................
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