1

INTERESTING CANCELLING

𝟏𝒏𝒔𝒊𝒏 𝒙=?

𝟏𝒏𝒔𝒊𝒏 𝒙=?

𝒔𝒊𝒙=𝟔

2

LIMITS OF TRIGONOMETRIC FUNCTIONS

In order to understand the derivatives that the trigonometric functions will produce, we must first understand how to evaluate two important trigonometric limits.

0

sinlimx

x

xThe first one is

3

The Sandwich Theorem

• First evaluate something that we know to be smaller

• Second evaluate something that we know to be larger.

• Make a conclusion about the value of the limit in between these small and large values.

4

Graph Y1 = sin x

x

x y =

-0.3 0.98507

-0.2 0.99335

-0.1 0.99833

0 undefined ÷ by 0

0.1 0.99833

0.2 0.99335

0.3 0.98507

10

First we will examine the value of for values of x close to 0.

sin x

x

sin x

x

We see in the table as x 0 1𝐬𝐢𝐧 𝒙𝒙

5

Graph Y1 = sin x

x

0

sinlim 1x

x

x

0

sinlim 1x

x

x

0

sinlim 1x

x

xSince and then

0

sinlim 1x

x

x

0

sinlim 1x

x

x

6

We need to review some trigonometry before we can proceed to the proof that

0

sinlimx

x

x

Slides 7 to 13 are included for those students interested in looking at the formal proof of this limit.

We will now move on to slide 14

7

The Circle

x

yr

x 2 + y 2 = r 2

r

ysin

r

xcos

x

ytan

8

Unit Circle

(0,-1)

(-1, 0) (1, 0)

(0, 1)x 2 + y 2 = 1

1r

r

cosr

x

sinr

y

(cos θ, sin θ)

9

Areas of Sectors in Degrees

Area of circle = p r2

If θ = 90o then the sector is or of the circle.o

o

360

904

1

10

Areas of Sectors in Radians

360o = 2p radians

11

(1, 0)(cos θ, 0)

(0, 1)

O

A

B D

C

cos θ, sin θ (0, sin θ)

r = cos θ

r = 1

The size of ∆OAB is between the areas of sector OCB and sector OAD

12

Area of sectorOCB

Area of∆OAB

Area of sectorOAD

< <

22

1rA 2

2

1rA bhA

2

1< <

< < 2cos2

1A sincos2

1A 212

1A

divide by ½

divide by q cosq

cos

1 2

A

cos

sincosA

cos

cos 2

A < <

2cosA sincosA 21A< <

cosAsin

A cos

1A< <

13

In order to evaluate our limit, we now need to look at what happens as θ→0

REMEMBER: cos 0o = 1

As we approach this limit from the left and from the right, it approaches the value of 1.

0 0 0

sin 1lim cos lim lim

cos

0

sin1 lim 1

Conclusion:

0

sinlim 1

14

Example 1: Estimate the limit by graphing

-0.3 0.7767

-0.2 0.8967

-0.1 0.97355

0 Undefined

0.1 0.97355

0.2 0.8967

0.3 0.7767

0

sin 4lim

4x

x

xx

0 1

0

sin 4lim

4x

x

x= 1

lim𝑥→ 0

sin 4 𝑥4 𝑥

15

lim𝑥→ 0

sin𝒏𝑥𝒏𝑥

=1

If the coefficients on the x are equal the limit value will be 1

lim𝑥→ 0

sin𝟐𝑥𝟐 𝑥

=1 lim𝑥→ 0

sin𝟕𝑥𝟕 𝑥

=1 lim𝑥→ 0

sin𝟏𝟎 𝑥𝟏𝟎 𝑥

=1

16

Example 2: Evaluate the limit

Solution: Multiply top and bottom by 2:

Separate into 2 limits:

Evaluate

lim𝑥→ 0

sin 2𝑥𝑥

=1

lim𝑥→ 0

sin 2𝑥𝑥

×𝟐𝟐=¿ lim

𝑥→0

𝟐 sin 2 𝑥𝟐𝑥

¿

lim𝑥→ 0

𝟐× lim𝑥→ 0

sin 2𝑥𝟐𝑥

(𝟐 ) (𝟏 )=𝟐

17

Example 3: Evaluate the limit

Solution: Multiply top and bottom by 3:

Separate into 2 limits:

Evaluate

lim𝑥→ 0

sin 3 𝑥4 𝑥

×𝟑𝟑=¿¿

𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧𝟑 𝒙𝟒 𝒙

lim𝑥→ 0

34

× lim𝑥→0

sin 3 𝑥3𝑥

(𝟑𝟒 )(𝟏 )=𝟑𝟒

lim𝑥→ 0

𝟑 sin 3𝑥4 (𝟑𝑥 )

18

-0.03 0.015

-0.02 0.01

-0.01 0.005

0 Undefined

0.01 -0.005

0.02 -0.01

0.03 -0.015

0 0

THE SECOND IMPORTANT TRIGONOMETRIC LIMIT

lim𝒙→𝟎

𝐜𝐨𝐬 𝒙−𝟏𝒙

We see in the table as x→0 →0𝐜𝐨𝐬𝒙−𝟏

𝒙

19

Mathematical Proof for

Multiply top and bottom by the conjugate cos x + 1

Pythagorean Identitysin 2 x + cos 2 x = 1 cos 2 x – 1 = – sin 2 x

lim𝒙→𝟎

𝐜𝐨𝐬 𝒙−𝟏𝒙

lim𝒙→𝟎

cos 𝑥−1𝑥

×cos 𝑥+1cos 𝑥+1

=¿¿ lim𝒙→𝟎

cos2𝑥−1𝑥 ( cos 𝑥+1 )

lim𝒙→𝟎

− sin2 𝑥𝑥 ( cos 𝑥+1 )

=¿¿lim𝒙→𝟎

sin 𝑥× (−sin 𝑥 )𝑥 (cos𝑥+1 )

lim𝒙→𝟎

sin 𝑥𝑥

×lim𝒙→𝟎

−sin 𝑥cos𝑥+1

=¿(1 )( −sin 0cos 0+1 )=¿(1 )( − 0

1+1 )=0

20

lim𝑥→ 0

cos2 𝑥−12 𝑥

𝐥𝐢𝐦𝒙→𝟎

𝐜𝐨𝐬𝟐𝟐 𝒙−𝟏𝟐 𝒙 (𝐜𝐨𝐬𝟐 𝒙+𝟏)

×𝐜𝐨𝐬𝟐 𝒙+𝟏𝐜𝐨𝐬𝟐 𝒙+𝟏

𝐥𝐢𝐦𝒙→𝟎

−𝐬𝐢𝐧𝟐𝟐 𝒙𝟐 𝒙 (𝐜𝐨𝐬𝟐 𝒙+𝟏)

𝐥𝐢𝐦𝒙→𝟎

−𝐬𝐢𝐧𝟐𝒙𝟐𝒙

× 𝐥𝐢𝐦𝒙 →𝟎

𝐬𝐢𝐧𝟐𝒙𝐜𝐨𝐬𝟐𝒙+𝟏

−𝟏×𝐬𝐢𝐧𝟎

𝐜𝐨𝐬𝟎+𝟏=𝟎

21

Example 4: Evaluate the limit 0

2cos 2lim

5x

x

x

0

2 cos 1lim

5x

x

x

0 0

2 cos 1lim lim

5x x

x

x

20 0

5

22

Example 5: Evaluate the limit

0

cos 2 1lim

4 2x

x x

x

0 0

cos 2 1lim lim

4 2x x

xx

x

00 0

4

0

cos 2lim

8x

x x x

x

23

Example 6: Evaluate the limit

20

1 coslimx

x

x

20

1 cos 1 coslim

1 cosx

x x

x x

2

20

1 coslim

1 cosx

x

x x

2

20 0

sin 1lim lim

1 cosx x

x

x x

0 0 0

sin sin 1lim lim lim

1 cosx x x

x x

x x x

1 1 11 1

1 cos 0 1 1 2

24

EXAMPLE 7: 0

tanlim

4x

x

x

0

sincoslim4x

xx

x

0

sinlim

4 cosx

x

x x

0 0

sin 1lim lim

4 cosx x

x

x x

0

11 lim

4 cos 0x

1 11

4 4

REMEMBER: sin

tancos

xx

x

Solution

25

ASSIGNMENT QUESTIONS

lim𝑥→ 0

sin 2𝑥𝑥

1.

 

 

2

26

2.lim𝑥→ 0

sin2 3 𝑥𝑥2

 

 

  

9

27

 

lim𝑥→ 0

sin 𝑥tan 𝑥3.

 

 

 

28

4. Use your calculator to estimate the value of the following limit.

lim𝑥→ 0

sin 6 𝑥sin 3 𝑥

x -0.2 -0.1 -0.01 0 0.01 0.1 0.2  1.651 1.911 1.999 ERR 1.999 1.911 1.651

2

0

2

29

Algebraic Method

lim𝑥→ 0

6 𝑥6 𝑥

sin 6 𝑥

3𝑥3𝑥

sin 3 𝑥

𝐥𝐢𝐦𝒙→𝟎

𝟔 𝒙𝟑 𝒙

×

𝐬𝐢𝐧𝟔 𝒙𝟔 𝒙

𝐬𝐢𝐧𝟑 𝒙𝟑 𝒙

𝐥𝐢𝐦𝒙→𝟎

𝟐× 𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧𝟔 𝒙𝟔 𝒙

÷ 𝐥𝐢𝐦𝒙 →𝟎

𝐬𝐢𝐧𝟑 𝒙𝟑 𝒙

=𝟐

30

0

cos 1lim

sinx

x

x

ASSIGNMENT QUESTIONS

5.

Multiply by the conjugate

cos 1

cos 1

x

x

0

cos 1 cos 1lim

sin cos 1x

x x

x x

2

0

cos 1lim

sin cos 1x

x

x x

Remember cos2 x + sin2 x = 1 so cos2 x – 1 = –sin2 x Substitute

2

0 0

sin sinlim lim

sin cos 1 cos 1

00

1 1

x x

x x

x x x

00

1 1

31

lim𝑥→ 0

1− cos2𝑥𝑥2

6.

𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧𝟐𝒙𝒙𝟐

𝟏×𝟏=𝟏