( ) ( ) ( ) 2 6A: Special Trig Limits! Math 400

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1 2 6A: Special Trig Limits Math 400 This section focuses entirely on the its of 2 specific trigonometric functions. The use of Theorem and the indeterminate cases of Theorem are all considered. a The it of the function f () as approaches a ( ) and f () is continuous in the neighborhood of = a from the left OR right side where there is a point at a, f (a) Theorem One: f () = f (a) if f (a) REALSAND f () is continuous in the neighborhood of = a In many cases the it of f () as approaches a can be found by the use of Theorem. Eample Eample 2 π 2 sin() 2 sin() is continuous in 2 the neighborhood of π 2 so π 3 cos(3) 6 cos(3) is continuous in 6 the neighborhood of π 3 so π 2 sin() 2 = sin π 2 2 π 2 ( ) ( ) π 3 cos(3) ( ( )) ( ) = cos 3 π 3 6 π 3 = π = cos( π ) = 0 2π 2π = 0 Eample 3 π cos() cos() is continuous in the neighborhood of π so π cos() = cos(π) π = 2 π Lecture 2 6A Page of Eitel

2 The Indeterminate Cases for trigonometric functions Many times the use of Theorem to find the it a f () g() produces f (a) g(a) = a Real Number This result allows us to state that the it as approaches a is f (a). Sometimes the use of Theorem g(a) f () to find the it produces a value of f(a) Nonzero number =. When this occurs the a g() g(a) zero function has a vertical asymptote at = a and the it of the function as approaches a from the left or right side of a vertical asymptote at = a is either + or. If h(a) k(a) is a non zero number zero then there is a "vertical asymptote" in the graph at = a and f () a g() = + or and f () a + = + or g() a f () g() = + or or DNE Lecture 2 6A Page 2 of Eitel

3 Sometimes the use of Theorem to find the it a f () g() produces a value of f(a) g(a) = 0 0 How do you determine the it given the zero case. We call this the Indeterminate case for zero Theorem. There are 3 special it problems in this chapter that involve this condition. sin and sin and + cos ( ) We will develop special theorems to find the it for each of there it problems Theorem sin = and sin = Lecture 2 6A Page 3 of Eitel

4 This it theorem requires that you have an eact match to the functions sin or and that sin the it be taken as. The theorem listed above cannot be used to answer its like the ones listed below. sin( 2 ) 2 sin( e ) e sin( ln() ) ln() Eample Eample 2 sin() 2 5 sin() = = 2 = 2 2 sin() sin() = = 5 = 5 5 sin() sin() = 2 = 5 Lecture 2 6A Page 4 of Eitel

5 Theorem If a then where a is a nonzero real number sin(a) a = and a = where a is a nonzero real number sina This it theorem requires that you have an eact match to the functions sin ( a) ) a or a sin( a) and that the it be taken as. The it of other functions cannot be taken using this theorem. Eample 3 Eample 4 sin(8) 8 3 sin(3) sin(a) a = for a nonzero real sin(a) a = for a nonzero real so so sin(8) 8 = 3 sin(3) = Eample 5 Eample 6 sin(4) 0 sin(7) multiply sin(4) by 4 4 multiply sin(7) by sin(4) 4 4 sin(4) sin(7) 7 7 sin(7) = 4 sin(4) sin(7) = 4 = 4 = 7 = 7 Lecture 2 6A Page 5 of Eitel

6 Eample 7 Eample 8 2sin(4) 3 3 sin 3 (2) rewrite as 2 3 sin(4) multiply 2 3 sin(4) sin(4) 4 sin(4) 4 by 4 4 rewrite as sin(2) sin(2) sin(2) multiply by sin(2) 2 sin(2) 2 sin(2) 2 sin(2) 2 sin(2) 2 sin(2) = 8 3 = 8 3 = 8 = 8 Lecture 2 6A Page 6 of Eitel

7 Theorem sin(a) a = a sin(a) = In English: You can multiply by and not change the it because = for all 0 The sin(a) a requires to approach BUT NOT REACH 0 so = as Eample 9 Eample 0 sin(4) 4 2 sin(2) multiply sin(4) 4 by multiply 2 sin(2) by sin(4) 4 2 sin(2) = sin(4) 4 = 2 sin(2) = 0 = 0 = DNE = DNE Lecture 2 6A Page 7 of Eitel

8 Eample Eample 2 sin(4 ) 2sin(5) cos(4) sin(3) rewrite as 2 sin(4) sin(5) rewrite as cos(4) sin(3) the sin(4) term needs a 4 in it's denominator and the sin(5) term needs a 5 in it's numarator multiply sin(4) by 4 4 and 5 by sin(5) 5 the cos(4) term needs a 4 in it's denominator and the sin(3) term needs a 3 in it's numarator multiply cos(4) by 4 4 and 3 by sin(3) 3 = 4 sin(4) sin(5) = 4 ( cos(4) ) sin(3) = sin(4) 4 sin(4) 4 = 4 cos(4) sin(3) = 2 5 sin(4) 4 sin(4) 4 = 4 cos(4) sin(3) = 2 5 = 2 5 = = 0 Lecture 2 6A Page 8 of Eitel

9 Eample 3 Eample 4 sin 2 (4) sin(4) sin(4) each of the 's in the denominator needs to be a 4 multiply by 6 6 = = 5 4sin(4) 4sin(4) 4 4 rewrite as cos(4) sin(3) cos(4) sin(3) the cos(4) needs a 4 in it's denominator and the term needs a 3 in it's numarator multiply by 3 4 = 6 5 sin(4) sin(4) 4 4 = cos(4) 4 3 sin(3) = 6 5 sin(4) 4 sin(4) 4 = cos(4) 4 3 sin(3) = 6 5 = 6 5 = 0 = 0 Lecture 2 6A Page 9 of Eitel

10 Equivalent infinitesimals When two functions f() and g() converge to zero at the same point and called equivalent infinitesimals. f() =, they are g() For the evaluation of the indeterminate form 0/0, we can use the following equivalent infinitesimals. sin arcsin π 2 arccos tan arc tan ln( +) e sin = arcsin = π 2 arccos = tan = arc tan ln( +) e = = = a ( +) a ( +) a a = 2 2 cos co 2 2 = Lecture 2 6A Page 0 of Eitel

11 Theorem cos( ) = 0 and cos( ) = 0 This it theorem requires that you have an eact match to the functions cos( ) and that the it be taken as. The it of other functions cannot be taken using this theorem. Note: You cannot use the theorem above to find the it of the reciprocal of + cos( ) = + cos( ) = Eample Eample 2 or cos ( ) cos( ) cos( ) 2 4 cos( ) 4 = = 2 = 2 0 = 0 2 cos ( ) cos( ) = = 4 = 4 0 = 0 4 cos( ) cos( ) Lecture 2 6A Page of Eitel

12 Take a unit circle and mark out the angle θ above and below the dashed radius, for a total angle of 2θ. The length of the arc of a unit circle is the measure of the arc angle (in radians), so the length of the purple arc is 2θ. The length of the straight red segment is 2 sin(θ). If we take a small enough segment of an arc, it closely resembles a straight line segment. As θ tends to zero, the length of the purple arc and the length of the red segment get closer, and the it is one. Lecture 2 6A Page 2 of Eitel

13 Prove that θ 0 cos(θ) θ = 0 Starting with the sine it proof above we can prove this cosine it algebraically. We begin by multiplying the it by cos(θ) + cos(θ) + to get θ 0 cos(θ) cos(θ) + i θ cos(θ) + Multiplying the binomials in the numerator gives θ 0 cos 2 (θ) θ cos(θ) + ( ) The Pythagorean identity, sin 2 θ + cos 2 θ =, can be converted to cos 2 θ = sin 2 θ and cos 2 can replace sin 2 the numerator to get θ 0 sin 2 (θ) θ cos(θ) + ( ) We can then break this epression into two parts, one of which is our sine it, and the other which can be evaluated by direct substitution: θ 0 sin(θ) θ i sin(θ) cos(θ) + ( ) = θ 0 sin(θ) θ i θ 0 sin(θ) cos(θ) + ( ) = i sin(0) cos(0) + ( ) = i 0 = 0 Lecture 2 6A Page 3 of Eitel

14 Using the figure below, we can also prove the cosine it geometrically, though it's not quite as obvious as the sine it. We start with a larger version of the unit circle (only a sector of angle 2θ is shown). Taking the radius as r = and subtracting cos(θ), which is the length of the green line, we have the length of the segment that spans the gap between the chord and the sector arc, - cos(θ). As angle θ shrinks (approaches zero), so does the length of the black segment. Lecture 2 6A Page 4 of Eitel