Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.5.77a
Capitolo 7, Problema 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

Guida verificata passo dopo passo
1
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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.
Video consigliato:
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.
Video consigliato:
05:21
Finding 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.
Video consigliato:
05:21
Finding Limits by Direct Substitution
Pratica correlata
Domanda del libro di testo

In Exercises 41–44:

a. Find f⁻¹(x).


41. f(x) = 2x + 3, a = −1

26
views
Domanda del libro di testo

Evaluate the integrals in Exercises 67–74 in terms of

a. inverse hyperbolic functions.

67. ∫(from 0 to 2√3)dx/√(4+x²)

7
views
Domanda del libro di testo

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

34
views
Domanda del libro di testo

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.

7. a. sec^(-1)(-√2)

26
views
Domanda del libro di testo

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

20
views
Domanda del libro di testo

21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for any positive integer n, even x^(1/1,000,000).

28
views