Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0^{+}}(\ln x \cot x)$$

Step-by-Step Solution

Answer

The limit is \(-\infty\).

Step 1: Check the Form

First, we need to establish that the limit is of the indeterminate form. We have the expression \( \ln(x) \cot(x) \). As \( x \to 0^+ \), \( \ln(x) \to -\infty \) and \( \cot(x) = \frac{\cos(x)}{\sin(x)} \to +\infty \). So, the expression is of the indeterminate form \( (-\infty) \times (+\infty) \). We need to convert it to a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form to apply l'Hôpital's Rule.

Step 2: Rewriting to Use l'Hôpital's Rule

The given expression \( \ln(x) \cot(x) \) can be rewritten as \( \frac{\ln(x)}{\tan(x)} \) to better apply l'Hôpital's Rule. Note that \( \tan(x) \to 0 \) as \( x \to 0^+ \), and \( \ln(x) \to -\infty \). This gives us the \( \frac{-\infty}{0} \), which further needs to be verified for l'Hôpital's Rule.

Step 3: Confirm Indeterminate Form for l'Hôpital's Rule

To use l'Hôpital's Rule, confirm \( \frac{-\infty}{0} \) as a suitable form. Rewrite \( \ln(x) \) and \( \tan(x) \) so that as \( x \to 0^+ \), it becomes \( \frac{-\infty}{0^+} \), which leads to \( \frac{-\infty}{\infty} \). This is technically indeterminate; instead, change to \( \frac{-\ln(x)}{\tan(x)} \). Both tend towards \( -\infty \) and \( \infty \) respectively, resulting in the form needed.

Step 4: Differentiate Numerator and Denominator

Applying l'Hôpital's Rule, differentiate the numerator \( f(x) = -\ln(x) \) and the denominator \( g(x) = \tan(x) \). We have \( f'(x) = -\frac{1}{x} \) and \( g'(x) = \sec^2(x) \). Now evaluate the limit: \( \lim_{x \to 0^+} \frac{-\frac{1}{x}}{\sec^2(x)} \).

Step 5: Simplify and Evaluate the Limit

Simplify the expression: \( \frac{-\frac{1}{x}}{\sec^2(x)} = -\frac{1}{x} \cdot \cos^2(x) \). As \( x \to 0^+ \), \( \cos^2(x) \to 1 \), hence the limit becomes \( \lim_{x \to 0^+} -\frac{1}{x} \), which approaches \( -\infty \). Thus, \( \lim_{x \rightarrow 0^{+}}(\ln x \cot x) = -\infty \).

Key Concepts

Indeterminate Forms

Indeterminate forms occur when evaluating limits results in expressions that do not have a clear value. The most common indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), but others, like \( 0 \cdot \infty \), \( \infty - \infty \), and \( 0^0 \), can also be seen. These forms arise frequently in calculus, especially when the behavior of a function around a certain point is not straightforward or predictable because the limit involves infinity.

• For example, \( 0 \cdot \infty \) suggests multiplying zero by an essentially undefined term, leading to ambiguity.
• In our original problem, \( \ln(x) \cdot \cot(x) \) becomes \( (-\infty) \times (+\infty) \) as \( x \to 0^+ \).

Understanding these forms is crucial since they signal us to use methods like l'Hôpital's Rule to simplify the limit evaluation. Such methods transform the original indeterminate form into something more manageable, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where we can apply derivatives to resolve the ambiguity.

Limits

Limits help us understand the behavior of functions as they approach a particular input. In simple terms, a limit describes the value a function gets closer to but doesn't necessarily ever reach as the input parameter approaches a specified value. Limits are foundational in calculus and necessary for defining derivatives and integrals.
In our problem, finding \( \lim_{x \to 0^+}(\ln x \cdot \cot x) \) involves determining what value the expression heads towards as \( x \) approaches zero from the positive side.

• As \( x \to 0^+ \), both \( \ln(x) \to -\infty \) and \( \cot(x) \to +\infty \), forming an indeterminate product.
• Rewriting as a fraction, \( \frac{\ln(x)}{\tan(x)} \) helps apply l'Hôpital's Rule. This conversion is essential to handle indeterminate forms effectively.

By examining limits, we obtain a deeper insight into the function's behavior, essential for solving complex calculus problems.

Differentiation

Differentiation is a fundamental concept in calculus involving computing a derivative, which represents the rate of change of a function with respect to one of its variables. By differentiating, we measure how changes in one variable affect the function, essential for solving rates and optimizing functions.
In our problem, after rewriting the product \( \ln(x) \cot(x) \) to \( \frac{\ln(x)}{\tan(x)} \), differentiation aids in applying l'Hôpital's Rule.

• We differentiate the numerator \( f(x) = -\ln(x) \), giving \( f'(x) = -\frac{1}{x} \).
• For the denominator \( g(x) = \tan(x) \), the derivative is \( g'(x) = \sec^2(x) \).

These derivatives are substituted back into the limit \( \lim_{x \to 0^+} \frac{-\frac{1}{x}}{\sec^2(x)} \). Through simplification, the expression simplifies further, revealing how the initial function behaves around the troubling point. Understanding differentiation and its applications is vital for anyone working with limits and continuity in calculus, as it provides the tools needed to interpret subtle behaviors of functions.