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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

L’Hôpital’s Rule

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 limits involving exponential, logarithmic, or trigonometric functions.
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Dominant Terms in Limits

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