Evaluate each limit. lim⁡x→π2sin⁡(x)−1sin⁡(x)−1{\displaystyle\lim_{x\to\frac{\pi}{2}}{\frac{\sin\left(x\right)-1}{\sqrt{\sin\left(x\right)}-1}}}

Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 2 - Limits

문제 2.6.55

2장, 문제 2.6.55

Evaluate each limit.

$\lim_{x \to \frac{π}{2}} \frac{\sin (x) - 1}{\sqrt{\sin (x)} - 1}$

검증된 단계별 안내

Identify the limit expression: $ \lim_{x \to \frac{\pi}{2}} \frac{\sin(x) - 1}{\sqrt{\sin(x)} - 1} $.

Substitute $ x = \frac{\pi}{2} $ into the expression to check for indeterminate form: $ \sin\left(\frac{\pi}{2}\right) = 1 $, leading to $ \frac{0}{0} $.

Apply L'Hôpital's Rule, which is used for limits of the form $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $, by differentiating the numerator and the denominator separately.

Differentiate the numerator: $ \frac{d}{dx}[\sin(x) - 1] = \cos(x) $.

Differentiate the denominator: $ \frac{d}{dx}[\sqrt{\sin(x)} - 1] = \frac{1}{2\sqrt{\sin(x)}} \cdot \cos(x) $.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. In this question, we are evaluating the limit of a function as x approaches π/2, which is crucial for understanding the behavior of the function near that point.

L'Hôpital's Rule

L'Hôpital's Rule is a method used to evaluate limits that result in indeterminate forms, such as 0/0 or ∞/∞. When faced with such forms, we can differentiate the numerator and denominator separately and then take the limit again, which can simplify the evaluation process.

Continuity and Discontinuity

Continuity refers to a function being unbroken and having no gaps at a point, while discontinuity indicates a break or jump in the function's value. Understanding whether the function is continuous at x = π/2 helps determine if the limit can be directly evaluated or if further analysis is needed.

Evaluate each limit. limx→π2sin(x)−1sin(x)−1{\(\displaystyle\[\lim\)_{x\(\to\]\frac{\pi}{2}\)}{\(\frac{\sin\left(x\right)-1}{\sqrt{\sin\left(x\right)}\)-1}}}