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U3 L2 Limits of Primary Trig Functions
UNIT 3 Lesson 2 Limits of Primary Trig Functions
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U3 L2 Limits of Primary Trig Functions
INTERESTING CANCELLING 1 π sin π₯=? 1 π sin π₯=? π ππ₯=6
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LIMITS OF TRIGONOMETRIC FUNCTIONS
U3 L2 Limits of Primary Trig Functions 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. The first one is π₯π’π¦ πβπ π¬π’π§ π π
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U3 L2 Limits of Primary Trig Functions
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.
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U3 L2 Limits of Primary Trig Functions
First we will examine the value of sin π₯ π₯ for values of x close to 0. Graph Y 1 = sin π₯ π₯ x Y 1 = sin π₯ π₯ β 0.3 β 0.2 β0.1 undefined Γ· by 0 0.1 0.2 0.3 1 We see in the table as x
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U3 L2 Limits of Primary Trig Functions
Graph Y 1 = sin π₯ π₯ Since and then
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U3 L2 Limits of Primary Trig Functions
We need to review some trigonometry before we can proceed to the proof that π₯π’π¦ πβπ π¬π’π§ π π Slides 7 to 14 are included for those students interested in looking at the formal proof of this limit. We will now move on to slide 15
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U3 L2 Limits of Primary Trig Functions
The Circle x y r x 2 + y 2 = r 2
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U3 L2 Limits of Primary Trig Functions
Unit Circle x 2 + y 2 = 1 (0,-1) (-1, 0) (1, 0) (0, 1) (cos ΞΈ, sin ΞΈ)
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Areas of Sectors in Degrees
U3 L2 Limits of Primary Trig Functions Areas of Sectors in Degrees Area of circle = p r2 If ΞΈ = 90o then the sector is or of the circle.
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U3 L2 Limits of Primary Trig Functions
Areas of Sectors in Radians 360o = 2p radians
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U3 L2 Limits of Primary Trig Functions
(1, 0) (cos ΞΈ, 0) (0, 1) O A B D C cos ΞΈ, sin ΞΈ (0, sin ΞΈ) r = 1 r = cos ΞΈ The size of βOAB is between the areas of sector OCB and sector OAD
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U3 L2 Limits of Primary Trig Functions
Area of sector OCB Area of βOAB OAD < < < divide by Β½ < divide by q cosq < <
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U3 L2 Limits of Primary Trig Functions
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. Conclusion:
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Example 1: Estimate the limit by graphing
U3 L2 Limits of Primary Trig Functions Example 1: Estimate the limit by graphing x -0.3 0.7767 -0.2 0.8967 -0.1 Undefined 0.1 0.2 0.3 1 = 1
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Example 2: Evaluate the limit
U3 L2 Limits of Primary Trig Functions Example 2: Evaluate the limit Solution: Multiply top and bottom by 2: Separate into 2 limits: Evaluate
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Example 3: Evaluate the limit
U3 L2 Limits of Primary Trig Functions Example 3: Evaluate the limit Solution: Multiply top and bottom by 3: Evaluate Separate into 2 limits:
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U3 L2 Limits of Primary Trig Functions
THE SECOND IMPORTANT TRIGONOMETRIC LIMIT -0.03 0.015 -0.02 0.01 -0.01 0.005 Undefined -0.005 0.02 0.03 -0.015 We see in the table as xβ0 β0
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U3 L2 Limits of Primary Trig Functions
Mathematical Proof for U3 L2 Limits of Primary Trig Functions π₯π’π¦ πβπ ππ¨π¬ πβπ π Multiply top and bottom by the conjugate cos x + 1 βlim π₯β0 sin π₯ π₯ lim π₯β0 sin π₯ cos π₯+1 lim π₯β0 cos π₯β1 π₯ Γ cos π₯+1 cos π₯+1 β1 sin (0) cos 0 +1 lim π₯β0 cos 2 π₯ β1 π₯( cos π₯+1) β =0 Pythagorean Identity sin 2 x + cos 2 x = 1 cos 2 x β 1 = β sin 2 x π₯π’π¦ πβπ ππ¨π¬ πβπ π =π lim π₯β0 βsin 2 π₯ π₯( cos π₯+1)
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Example 3: Evaluate the limit π₯π’π¦ πβπ ππ¨π¬ ππβπ ππ
π₯π’π¦ πβπ ππ¨π¬ ππβπ ππ lim π₯β0 cos 2π₯β1 2π₯ Γ ππ¨π¬ ππ+π ππ¨π¬ ππ+π π₯π’π¦ πβπ ππ¨π¬ π ππ βπ ππ (ππ¨π¬ ππ+π) π¬π’π§ π ππ+ ππ¨π¬ π ππ =π ππ¨π¬ π ππ =βπ¬π’π§ π ππ+π π₯π’π¦ πβπ β π¬π’π§ π ππ ππ (ππ¨π¬ ππ+π) ππ¨π¬ π ππβπ =βπ¬π’π§ π ππ π₯π’π¦ πβπ β π¬π’π§ ππ ππ Γ π₯π’π¦ πβπ π¬π’π§ ππ ππ¨π¬ ππ+π βπΓ π¬π’π§ π(π) ππ¨π¬ π(π)+π βπΓπ=π
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Example 4: Evaluate the limit
U3 L2 Limits of Primary Trig Functions Example 4: Evaluate the limit π₯π’π¦ πβπ π ππ¨π¬ πβπ ππ π₯π’π¦ πβπ π (ππ¨π¬ πβπ) ππ π₯π’π¦ πβπ π π Γ π₯π’π¦ πβπ ππ¨π¬ πβπ π π π Γπ=π
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U3 L2 Limits of Primary Trig Functions
Example 5: Evaluate the limit π₯π’π¦ πβπ π ππ¨π¬ ππβπ ππ π₯π’π¦ πβπ π (ππ¨π¬ ππβπ) π(ππ) π₯π’π¦ πβπ π π Γ π₯π’π¦ πβπ ππ¨π¬ ππβπ ππ π π Γπ=π
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U3 L2 Limits of Primary Trig Functions
Example 6: Evaluate the limit π₯π’π¦ πβπ πβ ππ¨π¬ π π π π₯π’π¦ πβπ πβ ππ¨π¬ π π π Γ π+ ππ¨π¬ π π+ ππ¨π¬ π π₯π’π¦ πβπ πβ ππ¨π¬ π π π π (π+ ππ¨π¬ π) π₯π’π¦ πβπ π¬π’π§ π π π π Γ π₯π’π¦ πβπ π π+ ππ¨π¬ π π π Γ π π+ ππ¨π¬ π = π π
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U3 L2 Limits of Primary Trig Functions
EXAMPLE 7: π₯π’π¦ πβπ πππ§ π ππ πππ§ π = π¬π’π§ π ππ¨π¬ π π₯π’π¦ πβπ π¬π’π§ π ππ¨π¬ π ππ π₯π’π¦ πβπ π¬π’π§ π ππ ππ¨π¬ π π₯π’π¦ πβπ π¬π’π§ π π Γ π₯π’π¦ πβπ π π ππ¨π¬ π πΓ π π( ππ¨π¬ π) = π π
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π₯π’π¦ πβπ π¬π’π§ ππ π Γ π π = π₯π’π¦ πβπ πΓ π₯π’π¦ πβπ π¬π’π§ ππ ππ = π
ASSIGNMENT QUESTIONS 1. π₯π’π¦ πβπ πππ ππ π π₯π’π¦ πβπ π¬π’π§ ππ π Γ π π = π₯π’π¦ πβπ πΓ π₯π’π¦ πβπ π¬π’π§ ππ ππ = π 2. πππ πβπ πππ π ππ π π π₯π’π¦ πβπ π¬π’π§ ππ π Γ π π Γ π₯π’π¦ πβπ π¬π’π§ ππ π Γ π π ππ₯π’π¦ πβπ π¬π’π§ ππ ππ Γπ π₯π’π¦ πβπ π¬π’π§ ππ ππ =π
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lim π₯β0 sin π₯ tan π₯ π₯π’π¦ πβπ π¬π’π§ π π¬π’π§ π ππ¨π¬ π
3. lim π₯β0 sin π₯ tan π₯ π₯π’π¦ πβπ π¬π’π§ π π¬π’π§ π ππ¨π¬ π π₯π’π¦ πβπ π¬π’π§ π Γ· π¬π’π§ π ππ¨π¬ π π₯π’π¦ πβπ π¬π’π§ π Γ ππ¨π¬ π π¬π’π§ π π₯π’π¦ πβπ ππ¨π¬ π ππ¨π¬ π=π
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4. Use your calculator to estimate the value of the following limit.
lim π₯β0 sin 6π₯ sin 3π₯ x -0.2 -0.1 -0.01 0.01 0.1 0.2 1.651 1.911 1.999 ERR 2 2
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Algebraic Method lim π₯β0 6π₯ 6π₯ sin 6π₯ 3π₯ 3π₯ sin 3π₯
π₯π’π¦ πβπ ππ ππ Γ π¬π’π§ ππ ππ π¬π’π§ ππ ππ π₯π’π¦ πβπ πΓ π₯π’π¦ πβπ π¬π’π§ ππ ππ Γ· π₯π’π¦ πβπ π¬π’π§ ππ ππ = π
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U3 L2 Limits of Primary Trig Functions
ASSIGNMENT QUESTIONS 5. Multiply by the conjugate Remember cos2 x + sin2 x = 1 so cos2 x β 1 = βsin2 x Substitute
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π₯π’π¦ πβπ πβ ππ¨π¬ π π π π = π₯π’π¦ πβπ π¬π’π§ π π π π =πΓπ=π
6. lim π₯β0 1β cos 2 π₯ π₯ 2 π₯π’π¦ πβπ πβ ππ¨π¬ π π π π = π₯π’π¦ πβπ π¬π’π§ π π π π =πΓπ=π
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