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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