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In each of Exercises \(69-76,\) use l'Hôpital's Rule to evaluate the one-sided limit. \(\lim _{x \rightarrow(\pi / 2)^{-}}(x-\pi / 2) \cot (x)\)

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify Indeterminate Form

First, observe that as \(x\) approaches \(\frac{\pi}{2}^-\), both \((x - \frac{\pi}{2})\) and \(\cot(x)\) become problematic. \(\cot(x) = \frac{\cos(x)}{\sin(x)}\) becomes undefined at \(\frac{\pi}{2}\), and \(x - \frac{\pi}{2}\) approaches zero. Therefore, the given expression is of the form \(0 \times \infty\). To apply l'Hôpital's Rule, we need to convert it to an \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.
02

Rewrite as a Fraction

Rewrite the expression \((x - \frac{\pi}{2}) \cot(x)\) as a fraction: \(\frac{x - \frac{\pi}{2}}{\tan(x)}\). Now the limit is \(\lim_{x \to (\pi/2)^-} \frac{x - \frac{\pi}{2}}{\tan(x)}\). Both the numerator and the denominator approach zero as \(x\) approaches \(\frac{\pi}{2}^-\), fitting the \(\frac{0}{0}\) indeterminate form necessary for l'Hôpital's Rule.
03

Apply l'Hôpital's Rule

Differentiate the numerator and the denominator with respect to \(x\). The derivative of \(x - \frac{\pi}{2}\) is 1. The derivative of \(\tan(x)\) is \(\sec^2(x)\), leading to the new limit: \(\lim_{x \to (\pi/2)^-} \frac{1}{\sec^2(x)}\).
04

Simplify the Expression

Since \(\sec(x) = \frac{1}{\cos(x)}\), \(\sec^2(x) = \frac{1}{\cos^2(x)}\). Therefore, \(\lim_{x \to (\pi/2)^-} \frac{1}{\sec^2(x)} = \lim_{x \to (\pi/2)^-} \cos^2(x)\).
05

Evaluate the Limit

As \(x\) approaches \(\frac{\pi}{2}^-\), \(\cos(x)\) approaches zero from the positive side. Therefore, \(\cos^2(x)\) approaches 0. Thus, the limit is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
Indeterminate forms are expressions encountered in calculus that do not initially produce a clear answer. When you evaluate a limit and get an expression like \(0 \times \infty, \frac{0}{0}, \frac{\infty}{\infty}, 0^0, \infty - \infty,\) or \(1^\infty\), these are classic examples of indeterminate forms. They require special techniques for evaluation.

In this exercise, as \(x\) approaches \(\frac{\pi}{2}^-\), the expression \((x - \frac{\pi}{2}) \cot(x)\) is initially in the form \(0 \times \infty\). Here, \(x - \frac{\pi}{2}\) tends towards zero, and \(\cot(x)\), which is \(\frac{\cos(x)}{\sin(x)}\), behaves like it tends towards infinity because \(\sin(x)\) approaches zero.

To make sense of it, we must convert it into a fraction. By rewriting it as \(\frac{x - \frac{\pi}{2}}{\tan(x)}\), it becomes eligible for l'Hôpital's Rule after recognizing it as a \(\frac{0}{0}\) form. Understanding these transformations is key to resolving indeterminate forms in calculus.
One-Sided Limits
One-sided limits focus on the behavior of functions as the input approaches a specific value from one side — either from the left or the right. Notationally, \(\lim_{x \to a^-}\) would mean approaching \(a\) from the left, and \(\lim_{x \to a^+}\) signifies approaching from the right.

In the given exercise, we examine the one-sided limit \(\lim_{x \to (\pi/2)^-}\). This means we’re looking at how the expression \((x - \frac{\pi}{2}) \cot(x)\) behaves as \(x\) gets closer to \(\frac{\pi}{2}\) from values smaller than \(\frac{\pi}{2}\).

Understanding one-sided limits can be crucial, especially when different behaviors arise from approaching from the left versus the right. They can unveil more about the function's nature than a straightforward two-sided limit, enlightening us on any discontinuities or asymmetries.
Calculus Limits
Understanding limits is fundamental in calculus as it captures how functions behave near specific points. Limits help in defining other core concepts like derivatives and integrals. They essentially detail what value a function approaches as the input gets closer to a particular point.

In this exercise, the limit expression is \(\lim_{x \to (\pi/2)^-} (x - \frac{\pi}{2}) \cot(x)\). Initially, this doesn't yield a definitive result when directly substituting \(x = \frac{\pi}{2}\). As such, calculus limits help us in deducing the behavior of the function's output near this crucial point.

Using rules like l'Hôpital's Rule, we can simplify the intricacies involved in resolving limits that result in indeterminate forms. The outcome here is a limit that approaches zero, suggesting a subtle trend in the function's behavior near \(\frac{\pi}{2}\). Understanding and calculating limits ensures that we can predict and comprehend the behavior of complex functions.
Trigonometric Limits
Trigonometric limits often involve functions like sine, cosine, and tangent and require understanding of their behaviors around key points. These functions have periodic and sometimes undefined values at certain points which may lead to indeterminate forms.

In the current problem, we deal with the limit of \((x - \frac{\pi}{2}) \cot(x)\) as \(x\) approaches \(\frac{\pi}{2}^-\). Since \(\cot(x) = \frac{\cos(x)}{\sin(x)}\), and \(\sin(x)\) tends to zero as \(x\) gets near \(\frac{\pi}{2}\), it's essential to manage these trigonometric functions wisely.

These types of limits are often managed by rewriting the expressions in meaningful ways such as applying identities like \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). Subsequently, l'Hôpital's Rule is applied to aid in computing the limit. Thorough understanding of these trigonometric properties ensures accurate calculation and comprehension when facing such intriguing calculus challenges.

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