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௫→ 𝑥ቇ Theorems 1 and 4 4 𝜋 2 − lim ௫→ 𝑥 = 𝜋 2 𝜋 2 = 0 Theorem 2 5 lim ௫→଴ cos(𝑥) Theorem 9 6 lim ௫→ cos ቀ 𝜋 2 − 𝑥ቁ = 1 lim ௫→ଶ௞గ cos(𝑥) = 1 where 𝑘 = 0 Let’s then try using cos ቀ − 𝑥ቁ = sin(𝑥) No. STATEMENT REASON 1 lim ௫→ cos ቀ 𝜋 2 − 𝑥ቁ Given 2 lim ௫→ sin(𝑥) cos ቀ 𝜋 2 − 𝑥ቁ = sin(𝑥) 3 sin ቀ 𝜋 2 Theorem 8 4 lim ௫→ cos ቀ 𝜋 2 − 𝑥ቁ = 1 lim ௫→ቀቂଶ௞ା ቃగቁ sin(𝑥) = 1 where 𝑘 = 0
60 MODULE 13 LIMITS OF INVERSE TRIGONOMETRIC FUNCTIONS LIMITS OF INVERSE TRIGONOMETRIC FUNCTIONS Inverse trigonometric functions have restricted domains, unlike trigonometric functions. lim ௫→ଵ arcsin(𝑥) = lim ௫→ଵ arccsc(𝑥) = 𝜋 2 lim ௫→ିଵ arcsin(𝑥) = lim ௫→ିଵ arccsc(𝑥) = − 𝜋 2 lim ௫→ଵ arccos(𝑥) = lim ௫→ଵ arcsec(𝑥) = 0 lim ௫→ିଵ arccos(𝑥) = lim ௫→ିଵ arcsec(𝑥) = 𝜋 lim ௫→ଵ arctan(𝑥) = lim ௫→ଵ arccot(𝑥) = 𝜋 4 lim ௫→ିଵ arctan(𝑥) = lim ௫→ିଵ arccot(𝑥) = − 𝜋 4 EVALUATING LIMITS OF TRIGONOMETRIC FUNCTIONS Evaluate: lim ௫→ିଶ arcsin ൬ 2 𝑥 There are two ways we could go about this: We can do direct substitution, or we can use arcsin ቀ ቁ = arccsc(𝑥) . Let’s use direct evaluation first. No. STATEMENT REASON 1 lim ௫→ିଶ arcsin ൬ 2 𝑥 Given 2 arcsin ൬ lim ௫→ିଶ 2 𝑥 Theorem 9 3 lim ௫→ିଶ 2 𝑥 = 2 lim ௫→ିଶ 1 𝑥 = 2 ⋅ 1 −2 = −1 Theorems 2, 3, and 6 4 lim ௫→ିଵ arcsin(𝑥) Theorem 9 5 lim ௫→ିଶ arcsin ൬ 2 𝑥 ൰ = − 𝜋 2 lim ௫→ିଵ arcsin(𝑥) = − 𝜋 2 Then let’s use the fact that arcsin ቀ ቁ = arccsc(𝑥) .
61 No. STATEMENT REASON 1 lim ௫→ିଶ arcsin ൬ 2 𝑥 Given 2 lim ௫→ିଶ arccsc ቀ 𝑥 2 arcsin ൬ 1 𝑥 ൰ = arccsc(𝑥) 3 lim ௫→ିଶ 𝑥 2 = 1 2 lim ௫→ିଶ 𝑥 = 1 2 ⋅ −2 ି = −1 ି Theorems 2 and 3 4 lim ௫→ିଵ arccsc(𝑥) Theorem 9 5 lim ௫→ିଶ arcsin ൬ 2 𝑥 ൰ = − 𝜋 2 lim ௫→ିଵ arccsc(𝑥) = − 𝜋 2 Now let’s evaluate: lim ௫→଴ arctan(𝑥 + 1) No. STATEMENT REASON 1 lim ௫→଴ arctan(𝑥 + 1) Given 2 arctan ቀlim ௫→଴ 𝑥 + 1ቁ Theorem 9 3 lim ௫→଴ 𝑥 + 1 = 0 + 1 = 1 Theorems 1, 2, and 4 4 lim ௫→ଵ arctan(𝑥) Theorem 9 5 lim ௫→଴ arctan(𝑥 + 1) = 𝜋 4 lim ௫→ଵ arctan(𝑥) = 𝜋 4
62 MODULE 14 LIMITS OF HYPERBOLIC FUNCTIONS LIMITS OF HYPERBOLIC FUNCTIONS As described in Module 3, hyperbolic functions are special types of exponential functions. lim ௫→୪୬൫√ଶ ାଵ൯ sinh(𝑥) = lim ௫→୪୬൫√ଶ ାଵ൯ csch(𝑥) = 1 lim ௫→୪୬൫√ଶ ିଵ൯ sinh(𝑥) = lim ௫→୪୬൫√ଶ ିଵ൯ csch(𝑥) = −1 lim ௫→଴ cosh(𝑥) = lim ௫→଴ sech(𝑥) = 1 lim ௫→଴ tanh(𝑥) = 0 EVALUATING LIMITS OF HYPERBOLIC FUNCTIONS Evaluate: lim ௫→ଵ cosh(1 − 𝑥) No. STATEMENT REASON 1 lim ௫→ଵ cosh(1 − 𝑥) Given 2 cosh ቀlim ௫→ଵ 1 − 𝑥ቁ Theorem 9 3 lim ௫→ଵ 1 − 𝑥 = 1 − lim ௫→ଵ 𝑥 = 1 − 1 = 0 Theorems 1, 2, and 4 4 lim ௫→଴ cosh(𝑥) Theorem 9 5 lim ௫→ଵ cosh(1 − 𝑥) = 1 lim ௫→଴ cosh(𝑥) = 1 Evaluate: lim ௫→୪୬൫ଷାଶ√ଶ sinh ቀ 𝑥 2 No. STATEMENT REASON 1 lim ௫→୪୬൫ଷାଶ√ଶ sinh ቀ 𝑥 2 Given 2 sinh ቆ lim ௫→୪୬൫ଷାଶ√ଶ 𝑥 2 Theorem 9 3 1 2 lim ௫→୪୬൫ଷାଶ√ଶ 𝑥 = 1 2 ln൫3 + 2√2 Theorems 2 and 3
63 No. STATEMENT REASON 4 𝑥 → 1 2 ln൫3 + 2√2 𝑥 → ln ቆ 3 + 2√2 𝑥 → ln ቆ 2 + 2√2 + 1 𝑥 → ln ቆ ൫√2 + 1൯ 𝑥 → ln൫√2 + 1൯ Simplify the logarithm first 5 lim ௫→୪୬൫√ଶ ାଵ൯ sinh(𝑥) Theorem 9 6 lim ௫→୪୬൫ଷାଶ√ଶ sinh ቀ 𝑥 2 ቁ = 1 lim ௫→୪୬൫√ଶ ାଵ൯ sinh(𝑥) = 1 Evaluate: lim ௫→ିଵ tanh(𝑥 + 1) No. STATEMENT REASON 1 lim ௫→ିଵ tanh(𝑥 + 1) Given 2 tanh ቀlim ௫→ଵ 𝑥 + 1ቁ Theorem 9 3 lim ௫→ିଵ 𝑥 + 1 = lim ௫→ିଵ 𝑥 + 1 = −1 + 1 = 0 Theorems 1, 2, and 4 4 lim ௫→଴ tanh(𝑥) Theorem 9 5 lim ௫→ିଵ tanh(𝑥 + 1) = 0 lim ௫→଴ tanh(𝑥) = 0

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