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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 2, Problem 2.25

Determine the following limits.


lim x→π/2 1/√sin x − 1 / x + π/2

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1
Identify the form of the limit as \( x \to \frac{\pi}{2} \). Notice that both terms in the expression \( \frac{1}{\sqrt{\sin x}} - \frac{1}{x + \frac{\pi}{2}} \) approach infinity, indicating an indeterminate form of type \( \infty - \infty \).
To resolve the indeterminate form, find a common denominator for the expression. The common denominator is \( \sqrt{\sin x} (x + \frac{\pi}{2}) \).
Rewrite the expression as a single fraction: \( \frac{(x + \frac{\pi}{2}) - \sqrt{\sin x}}{\sqrt{\sin x} (x + \frac{\pi}{2})} \).
Apply L'Hôpital's Rule if necessary, which involves differentiating the numerator and the denominator separately, since the limit is still in an indeterminate form.
Evaluate the limit of the resulting expression as \( x \to \frac{\pi}{2} \) after simplification.

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Key Concepts

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

Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. They are essential for understanding continuity, derivatives, and integrals. In this question, evaluating the limit as x approaches π/2 involves analyzing the behavior of the function near that point.
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Indeterminate Forms

Indeterminate forms occur when direct substitution in a limit leads to expressions like 0/0 or ∞/∞, which do not provide clear information about the limit's value. In this case, substituting x = π/2 into the limit expression results in an indeterminate form, necessitating further analysis, such as algebraic manipulation or L'Hôpital's Rule.
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L'Hôpital's Rule

L'Hôpital's Rule is a method used to evaluate limits that result in indeterminate forms. It states that if the limit of f(x)/g(x) results in 0/0 or ∞/∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator. This rule simplifies the process of finding limits, especially in complex expressions like the one presented in the question.
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