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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
Capitolo 7, Problema 7.5.19

Use l’Hôpital’s rule to find the limits in Exercises 7–52.


19. lim (θ → π/6) (sin θ - 1/2) / (θ - π/6)

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Identify the limit expression: \(\lim_{\theta \to \frac{\pi}{6}} \frac{\sin \theta - \frac{1}{2}}{\theta - \frac{\pi}{6}}\).
Check if the limit is an indeterminate form by substituting \(\theta = \frac{\pi}{6}\): since \(\sin \frac{\pi}{6} = \frac{1}{2}\), the numerator becomes \(\frac{1}{2} - \frac{1}{2} = 0\) and the denominator is \(\frac{\pi}{6} - \frac{\pi}{6} = 0\), so the limit is of the form \(\frac{0}{0}\).
Since the limit is an indeterminate form \(\frac{0}{0}\), apply l’Hôpital’s Rule by differentiating the numerator and denominator separately with respect to \(\theta\).
Compute the derivative of the numerator: \(\frac{d}{d\theta} (\sin \theta - \frac{1}{2}) = \cos \theta\), and the derivative of the denominator: \(\frac{d}{d\theta} (\theta - \frac{\pi}{6}) = 1\).
Rewrite the limit using these derivatives: \(\lim_{\theta \to \frac{\pi}{6}} \frac{\cos \theta}{1}\), then evaluate this limit by substituting \(\theta = \frac{\pi}{6}\).

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