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?

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