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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 7.5.78a
Capitolo 7, Problema 7.5.78a

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

Guida verificata passo dopo passo
1
Step 1: Identify the original limit expression: \(\lim_{x \to 0} \frac{x^{2} - 2x}{x^{2} - \sin x}\).
Step 2: Check if the limit can be directly substituted by plugging in \(x = 0\). If direct substitution leads to an indeterminate form like \(\frac{0}{0}\), then apply algebraic manipulation or L'Hôpital's Rule.
Step 3: The solution attempts to apply L'Hôpital's Rule by differentiating numerator and denominator separately. Differentiate numerator: \(\frac{d}{dx}(x^{2} - 2x) = 2x - 2\). Differentiate denominator: \(\frac{d}{dx}(x^{2} - \sin x) = 2x - \cos x\).
Step 4: Evaluate the new limit: \(\lim_{x \to 0} \frac{2x - 2}{2x - \cos x}\). Check if direct substitution is valid here. If it still results in an indeterminate form, consider applying L'Hôpital's Rule again or re-examining the differentiation.
Step 5: The next step in the problem shows a different limit: \(\lim_{x \to 0} \frac{2}{2 + \sin x}\), which does not follow from the previous step correctly. Verify the correctness of each differentiation and algebraic step to determine which parts are valid and which are not.

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Concetti chiave

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