University:
California State University, Northridge
Course:
MATH 250 | Calculus III
Academic year:

2023
Views:

111

Pages:

40
Author:

Aidan G.

Trigonometric Limits more examples of limits Substitution Theorem for Trigonometric Functions laws for evaluating limits Theorem A. For each point c in function’s domain: lim sin x = sin c, x→c lim tan x = tan c, x→c lim csc x = csc c, x→c lim cos x = cos c, x→c lim cot x = cot c, x→c lim sec x = sec c. x→c Theorem A. For each point c in function’s domain: lim sin x = sin c, x→c lim tan x = tan c, x→c lim csc x = csc c, x→c lim cos x = cos c, x→c lim cot x = cot c, x→c lim sec x = sec c. x→c Proof. Prove first that lim sin x = 0, x→0 lim cos x = 1. x→0 Is it obvious? lim sin x = 0, x→0 y=sin(x) lim cos x = 1. x→0 y=cos(x) Is it obvious? lim sin x = 0, x→0 lim cos x = 1. x→0 y=sin(x) No. The picture is not precise. y=cos(x) Is it obvious? lim sin x = 0, x→0 lim cos x = 1. x→0 y=sin(x) y=cos(x) No. The picture is not precise. Use definitions of sin(x) and cos(x). Use The One-Sided Squeeze Theorem. If f (x) ≤ g(x) ≤ h(x) near c and lim f (x) = x→c+ lim h(x) = Lright, then x→c+ y=h(x) L y=g(x) y=f(x) lim g(x) = Lright c x→c+ Also, if lim f (x) = lim h(x) = Lleft, then x→c− x→c− y=h(x) y=g(x) L lim g(x) = Lleft x→c− y=f(x) c Use The One-Sided Limits. lim g(x) = L x→c ⇔ lim g(x) = lim g(x) = L x→c− x→c+ And other Limits Theorems. lim[f (x) + g(x)] = lim f (x) + lim g(x), x→c x→c x→c lim[f (x)g(x)] = [lim f (x)][lim g(x)], x→c x→c x→c e.t.c. A 0 < AB < AC < arcAC t>0 O t<0 or, √ p 0 < sin(t) < 2 1 − cos(t) < t cos(t) t % sin(t) 1 t) os&(& c& & 1 & & 2% An estimate from geometry: B C A 0 < AB < AC < arcAC t>0 O cos(t) B t<0 or, √ p 0 < sin(t) < 2 1 − cos(t) < t By the Right-Sided Squeeze Theorem lim sin(x) = 0, x→0+ lim (1 − cos(x)) = 0, x→0+ t % sin(t) 1 t) os&(& c& & 1 & & 2% An estimate from geometry: C Similarly, lim sin(x) = 0, x→0− lim (1 − cos(x)) = 0. x→0− Similarly, lim sin(x) = 0, x→0− lim (1 − cos(x)) = 0. x→0− The left and the right limits are equal, thus lim sin(x) = 0, x→0 lim(1 − cos(x)) = 0 x→0 Similarly, lim sin(x) = 0, x→0− lim (1 − cos(x)) = 0. x→0− The left and the right limits are equal, thus lim sin(x) = 0, x→0 lim(1 − cos(x)) = 0 x→0 or, lim sin(x) = 0, x→0 lim cos(x) = 1. x→0 EXAMPLES EXAMPLE 1. Evaluate limit lim θ tan(θ) θ→π/4 Since θ = π/4 is in the domain of the function θ tan(θ) EXAMPLE 1. Evaluate limit lim θ tan(θ) θ→π/4 Since θ = π/4 is in the domain of the function θ tan(θ) we use Substitution Theorem to substitute π/4 for θ in the limit expression: π π π π lim θ tan θ = tan = ·1= . 4 4 4 4 θ→π/4 EXAMPLE 2. Evaluate limit cos2(θ) lim . θ→π/2 1 − sin(θ) Since at θ = π/2 the denominator of cos2(θ)/(1 − sin(θ)) turns to zero, we can not substitute π/2 for θ immediately. EXAMPLE 2. Evaluate limit cos2(θ) lim . θ→π/2 1 − sin(θ) Since at θ = π/2 the denominator of cos2(θ)/(1 − sin(θ)) turns to zero, we can not substitute π/2 for θ immediately. Instead, we rewrite the expression 2 using sin (θ) + cos2(θ) = 1: 2 1 − sin (θ) (1 − sin(θ))(1 + sin(θ)) lim = lim (1 − sin(θ)) θ→π/2 1 − sin(θ) θ→π/2 Finally, (1 − sin(θ))(1 + sin(θ)) lim (1 − sin(θ)) θ→π/2 Finally, (1 − sin(θ))(1 + sin(θ)) lim (1 − sin(θ)) θ→π/2 1 − sin(θ) = lim lim (1 + sin(θ)) θ→π/2 1 − sin(θ) θ→π/2 Finally, (1 − sin(θ))(1 + sin(θ)) lim (1 − sin(θ)) θ→π/2 1 − sin(θ) = lim lim (1 + sin(θ)) θ→π/2 1 − sin(θ) θ→π/2 = 1 · (1 + sin(π/2)) = 2. Special Trigonometric Limits sin(x)/x →? as x → 0 Theorem B1. lim x→0 sin x x = 1. Theorem B2. lim x→0 1 − cos x x = 0. Proof B1. A fact from geometry: (t > 0) area(OAB)≤area(ODB)≤area(ODC) cos2(t)t/2 ≤ sin(t) cos(t)/2 ≤ t/2. dividing by cos(t)t/2 get cos t ≤ sin t t ≤ 1 cos t Right-Sided Squeeze Theorem: lim t→0+ sin t t =1 The same inequality holds for t < 0: cos t ≤ sin t t ≤ 1 cos t Left-Sided Squeeze Theorem: lim t→0− sin t t =1 The left and the right limits are equal, thus, lim t→0 sin t t =1 Proof B2. By multiplying numerator and denominator with (1 + cos x) lim x→0 1 − cos x x = lim x→0 (1 − cos x) (1 + cos x) x (1 + cos x) Proof B2. By multiplying numerator and denominator with (1 + cos x) lim x→0 1 − cos x x = lim (1 − cos x) (1 + cos x) x→0 x 2 2 = lim x→0 (1 − cos x) x(1 + cos x) (1 + cos x) = lim x→0 sin x x(1 + cos x) Proof B2. By multiplying numerator and denominator with (1 + cos x) lim x→0 1 − cos x x = lim (1 − cos x) (1 + cos x) x→0 x 2 2 = lim (1 − cos x) (1 + cos x) = lim sin x x→0 x(1 + cos x) x(1 + cos x) Using B1 write sin x limx→0[sin x] = lim = 0. x→0 x limx→0[1 + cos x] x→0 EXAMPLES EXAMPLE 3. Evaluate limit tan t lim t→0 t EXAMPLE 3. Evaluate limit tan t lim t→0 t Recalling tan t = sin t/ cos t, and using B1: sin t = lim t→0 (cos t)t EXAMPLE 3. Evaluate limit tan t lim t→0 t Recalling tan t = sin t/ cos t, and using B1: h i sin t sin t 1 = lim = lim lim t→0 (cos t)t t→0 t t→0 cos t EXAMPLE 3. Evaluate limit tan t lim t→0 t Recalling tan t = sin t/ cos t, and using B1: h i sin t sin t 1 1 = lim = lim =1· =1 lim t→0 (cos t)t t→0 t t→0 cos t 1 EXAMPLE 4. Evaluate limit (Can’t use B1 !): sin(3t) lim t→0 t EXAMPLE 4. Evaluate limit (Can’t use B1 !): sin(3t) lim t→0 t multiply both numerator and denominator with 3: sin(3t) = lim 3 t→0 3t EXAMPLE 4. Evaluate limit (Can’t use B1 !): sin(3t) lim t→0 t multiply both numerator and denominator with 3: sin(3t) = lim 3 t→0 3t Now, t → 0 as 3t → 0, so sin(3t) sin x = lim 3 = 3. lim =1 x→0 x 3t→0 3t B1 applies (with a substitution x = 3t). EXAMPLE 5. Evaluate limit 1 − cos t lim t→0 sin t EXAMPLE 5. Evaluate limit 1 − cos t lim t→0 sin t Divide both numerator and denominator with t: 1 − cos t t = lim sin t t→0 t EXAMPLE 5. Evaluate limit 1 − cos t lim t→0 sin t Divide both numerator and denominator with t: 1 − cos t t = lim sin t t→0 t Use B1 and B2: 0 = = 0. 1