Upload
vernon-simson
View
218
Download
1
Embed Size (px)
Citation preview
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