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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 7.5.22

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


22. lim (x → 1) (x - 1) / (ln x - sin πx)

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First, verify that the limit is an indeterminate form of type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by substituting \( x = 1 \) into the expression \( \frac{x - 1}{\ln x - \sin \pi x} \).
Since direct substitution gives \( \frac{0}{0} \), apply l'Hôpital's Rule, which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) if the original limit is an indeterminate form.
Find the derivative of the numerator: \( f(x) = x - 1 \), so \( f'(x) = 1 \).
Find the derivative of the denominator: \( g(x) = \ln x - \sin \pi x \). Use the derivatives \( \frac{d}{dx} \ln x = \frac{1}{x} \) and \( \frac{d}{dx} \sin \pi x = \pi \cos \pi x \), so \( g'(x) = \frac{1}{x} - \pi \cos \pi x \).
Evaluate the new limit \( \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{1}{\frac{1}{x} - \pi \cos \pi x} \) by substituting \( x = 1 \).

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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.
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Limits Involving Logarithmic and Trigonometric Functions

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Derivative Computation

Calculating derivatives of functions such as (x - 1), ln x, and sin πx accurately is essential when applying l’Hôpital’s Rule. Knowing the derivative rules for polynomials, logarithms, and trigonometric functions ensures correct application of the rule.
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