1 interesting cancelling. 2 limits of trigonometric functions in order to understand the derivatives...
TRANSCRIPT
1
INTERESTING CANCELLING
πππππ π=?
πππππ π=?
πππ=π
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
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The Sandwich Theorem
β’ First evaluate something that we know to be smaller
β’ Second evaluate something that we know to be larger.
β’ Make a conclusion about the value of the limit in between these small and large values.
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Graph Y1 = sin x
x
x y =
-0.3 0.98507
-0.2 0.99335
-0.1 0.99833
0 undefined Γ· by 0
0.1 0.99833
0.2 0.99335
0.3 0.98507
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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 1π¬π’π§ ππ
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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
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The Circle
x
yr
x 2 + y 2 = r 2
r
ysin
r
xcos
x
ytan
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Unit Circle
(0,-1)
(-1, 0) (1, 0)
(0, 1)x 2 + y 2 = 1
1r
r
cosr
x
sinr
y
(cos ΞΈ, sin ΞΈ)
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Areas of Sectors in Degrees
Area of circle = p r2
If ΞΈ = 90o then the sector is or of the circle.o
o
360
904
1
10
Areas of Sectors in Radians
360o = 2p radians
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(1, 0)(cos ΞΈ, 0)
(0, 1)
O
A
B D
C
cos ΞΈ, sin ΞΈ (0, sin ΞΈ)
r = cos ΞΈ
r = 1
The size of βOAB is between the areas of sector OCB and sector OAD
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Area of sectorOCB
Area ofβOAB
Area of sectorOAD
< <
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1rA 2
2
1rA bhA
2
1< <
< < 2cos2
1A sincos2
1A 212
1A
divide by Β½
divide by q cosq
cos
1 2
A
cos
sincosA
cos
cos 2
A < <
2cosA sincosA 21A< <
cosAsin
A cos
1A< <
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In order to evaluate our limit, we now need to look at what happens as ΞΈβ0
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
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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
limπ₯β 0
sin 4 π₯4 π₯
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limπ₯β 0
sinππ₯ππ₯
=1
If the coefficients on the x are equal the limit value will be 1
limπ₯β 0
sinππ₯π π₯
=1 limπ₯β 0
sinππ₯π π₯
=1 limπ₯β 0
sinππ π₯ππ π₯
=1
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Example 2: Evaluate the limit
Solution: Multiply top and bottom by 2:
Separate into 2 limits:
Evaluate
limπ₯β 0
sin 2π₯π₯
=1
limπ₯β 0
sin 2π₯π₯
Γππ=ΒΏ lim
π₯β0
π sin 2 π₯ππ₯
ΒΏ
limπ₯β 0
πΓ limπ₯β 0
sin 2π₯ππ₯
(π ) (π )=π
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Example 3: Evaluate the limit
Solution: Multiply top and bottom by 3:
Separate into 2 limits:
Evaluate
limπ₯β 0
sin 3 π₯4 π₯
Γππ=ΒΏΒΏ
π₯π’π¦πβπ
π¬π’π§π ππ π
limπ₯β 0
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Γ limπ₯β0
sin 3 π₯3π₯
(ππ )(π )=ππ
limπ₯β 0
π sin 3π₯4 (ππ₯ )
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-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
limπβπ
ππ¨π¬ πβππ
We see in the table as xβ0 β0ππ¨π¬πβπ
π
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Mathematical Proof for
Multiply top and bottom by the conjugate cos x + 1
Pythagorean Identitysin 2 x + cos 2 x = 1 cos 2 x β 1 = β sin 2 x
limπβπ
ππ¨π¬ πβππ
limπβπ
cos π₯β1π₯
Γcos π₯+1cos π₯+1
=ΒΏΒΏ limπβπ
cos2π₯β1π₯ ( cos π₯+1 )
limπβπ
β sin2 π₯π₯ ( cos π₯+1 )
=ΒΏΒΏlimπβπ
sin π₯Γ (βsin π₯ )π₯ (cosπ₯+1 )
limπβπ
sin π₯π₯
Γlimπβπ
βsin π₯cosπ₯+1
=ΒΏ(1 )( βsin 0cos 0+1 )=ΒΏ(1 )( β 0
1+1 )=0
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limπ₯β 0
cos2 π₯β12 π₯
π₯π’π¦πβπ
ππ¨π¬ππ πβππ π (ππ¨π¬π π+π)
Γππ¨π¬π π+πππ¨π¬π π+π
π₯π’π¦πβπ
βπ¬π’π§ππ ππ π (ππ¨π¬π π+π)
π₯π’π¦πβπ
βπ¬π’π§ππππ
Γ π₯π’π¦π βπ
π¬π’π§ππππ¨π¬ππ+π
βπΓπ¬π’π§π
ππ¨π¬π+π=π
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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
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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
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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
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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
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ASSIGNMENT QUESTIONS
limπ₯β 0
sin 2π₯π₯
1.
2
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2.limπ₯β 0
sin2 3 π₯π₯2
9
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limπ₯β 0
sin π₯tan π₯3.
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4. Use your calculator to estimate the value of the following limit.
limπ₯β 0
sin 6 π₯sin 3 π₯
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
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Algebraic Method
limπ₯β 0
6 π₯6 π₯
sin 6 π₯
3π₯3π₯
sin 3 π₯
π₯π’π¦πβπ
π ππ π
Γ
π¬π’π§π ππ π
π¬π’π§π ππ π
π₯π’π¦πβπ
πΓ π₯π’π¦πβπ
π¬π’π§π ππ π
Γ· π₯π’π¦π βπ
π¬π’π§π ππ π
=π
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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 β 1 = βsin2 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
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limπ₯β 0
1β cos2π₯π₯2
6.
π₯π’π¦πβπ
π¬π’π§ππππ
πΓπ=π