Textbook Question
Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
67. ∫(from 0 to 2√3)dx/√(4+x²)
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.5.77a
77. Which one is correct, and which one is wrong? Give reasons for your answers.
a. lim (x → 3) (x - 3) / (x² - 3) = lim (x → 3) 1 / (2x) = 1/6
Verified step by step guidance1
Step 1: Identify the original limit expression: \(\lim_{x \to 3} \frac{x - 3}{x^{2} - 3}\).
Step 2: Check if direct substitution is possible by plugging \(x = 3\) into the expression: numerator becomes \(3 - 3 = 0\), denominator becomes \(3^{2} - 3 = 9 - 3 = 6\), so the expression evaluates to \(\frac{0}{6} = 0\).
Step 3: Notice that the original limit expression simplifies to \(\frac{0}{6} = 0\), so the limit should be 0, not \(\frac{1}{6}\).
Step 4: Analyze the second expression given: \(\lim_{x \to 3} \frac{1}{2x} = \frac{1}{6}\). This limit is correct for the function \(\frac{1}{2x}\), but it is not equivalent to the original limit expression.
Step 5: Conclude that the first limit expression is correct and equals 0, while the second expression is unrelated to the original limit and thus incorrect in this context.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limit of a Function
The limit of a function as x approaches a value describes the behavior of the function near that point. It helps determine the value the function approaches, even if it is not defined at that point. Understanding limits is essential for evaluating expressions that may initially appear undefined.
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06:11
Limits of Rational Functions: Denominator = 0
Direct Substitution and Indeterminate Forms
Direct substitution involves plugging the approaching value into the function to find the limit. However, if this results in an indeterminate form like 0/0, further algebraic manipulation or techniques like factoring or L'Hôpital's Rule are needed to evaluate the limit correctly.
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05:21Finding Limits by Direct Substitution
Algebraic Simplification in Limits
Simplifying expressions algebraically before taking limits can resolve indeterminate forms. For example, factoring numerator and denominator or canceling common terms can reveal the true behavior of the function near the point, enabling accurate limit evaluation.
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05:21Finding Limits by Direct Substitution
Related Practice
Textbook Question
Suppose that the function g and its derivative with respect to x have the following values at x=0, 1, 2, 3, and 4.
Assuming the inverse function g^(-1) is differentiable, find the slope of g^(-1)(x) at
a. x=1
Textbook Question
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
7. a. sec^(-1)(-√2)
Textbook Question
In Exercises 41–44:
a. Find f⁻¹(x).
42. f(x) = (x + 2) / (1 − x), a = 1/2
Textbook Question
78. Which one is correct, and which one is wrong? Give reasons for your answers.
a. lim (x → 0) (x² - 2x) / (x² - sin x) = lim (x → 0) (2x - 2) / (2x - cos x) = lim (x → 0) 2 / (2 + sin x) = 2 / (2 + 0) = 1
1
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Textbook Question
[Technology Exercise] In Exercises 139–141, find the domain and range of each composite function. Then graph the compositions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see.
139. a. y=arctan(tan x)