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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 7.6.108

L’Hôpital’s Rule
Find the limits in Exercises 103–110.
108. lim(x→∞)(e^x arctan(e^x))/(e^(2x)+x)

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Identify the limit expression: \(\lim_{x \to \infty} \frac{e^{x} \arctan(e^{x})}{e^{2x} + x}\).
Check the behavior of numerator and denominator as \(x \to \infty\): \(e^{x} \to \infty\), \(\arctan(e^{x}) \to \frac{\pi}{2}\), so numerator behaves like \(\infty \cdot \frac{\pi}{2} = \infty\). Denominator \(e^{2x} + x \to \infty\). So the limit is of the form \(\frac{\infty}{\infty}\), which is an indeterminate form suitable for L’Hôpital’s Rule.
Apply L’Hôpital’s Rule by differentiating numerator and denominator separately with respect to \(x\):
Numerator derivative: Use product rule on \(e^{x} \arctan(e^{x})\):
\[\frac{d}{dx} \left(e^{x} \arctan(e^{x})\right) = e^{x} \arctan(e^{x}) + e^{x} \cdot \frac{1}{1 + (e^{x})^{2}} \cdot e^{x} = e^{x} \arctan(e^{x}) + \frac{e^{2x}}{1 + e^{2x}}.\]
Denominator derivative: \(\frac{d}{dx} (e^{2x} + x) = 2 e^{2x} + 1\).

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