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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 7.5.24

Use l’Hôpital’s rule to find the limits in Exercises 7–52.
24. lim (x → π/2) (ln(csc x)) / (x - (π/2))²

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First, identify the form of the limit as \(x\) approaches \(\frac{\pi}{2}\). Evaluate the numerator and denominator separately at \(x = \frac{\pi}{2}\) to check if it results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
Rewrite the limit expression: \(\lim_{x \to \frac{\pi}{2}} \frac{\ln(\csc x)}{\left(x - \frac{\pi}{2}\right)^2}\). Note that \(\csc x = \frac{1}{\sin x}\), so \(\ln(\csc x) = \ln\left(\frac{1}{\sin x}\right) = -\ln(\sin x)\).
Since the denominator is squared, the limit is of the form \(\frac{0}{0}\) after substitution, so l’Hôpital’s Rule applies. Differentiate the numerator and denominator with respect to \(x\):
Calculate the derivative of the numerator: \(\frac{d}{dx} \left( \ln(\csc x) \right) = \frac{d}{dx} \left( -\ln(\sin x) \right) = -\frac{\cos x}{\sin x} = -\cot x\).
Calculate the derivative of the denominator: \(\frac{d}{dx} \left( x - \frac{\pi}{2} \right)^2 = 2 \left( x - \frac{\pi}{2} \right)\). Then, rewrite the limit as \(\lim_{x \to \frac{\pi}{2}} \frac{-\cot x}{2 \left( x - \frac{\pi}{2} \right)}\). If this still results in an indeterminate form, apply l’Hôpital’s Rule again by differentiating numerator and denominator once more.

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l’Hôpital’s Rule is a method for evaluating limits that result in indeterminate forms like 0/0 or ∞/∞. It states that the limit of a ratio of functions can be found by taking the limit of the ratio of their derivatives, provided certain conditions are met. This rule simplifies complex limit problems by transforming them into easier derivative calculations.
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