Problem 64 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow 0^{+}} x(\ln x)^{2} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Analyze the Function

First, let us understand the behavior of the function \( x(\ln x)^{2} \) as \( x \rightarrow 0^{+} \). Here, \( x(\ln x)^{2} \) implies two components: the linear term \( x \) and the logarithmic term \( (\ln x)^{2} \). As \( x \) approaches \( 0^{+} \), \( x \) goes to zero, and \( \ln x \) tends to \( -\infty \). Thus, \( (\ln x)^{2} \) approaches \( +\infty \). This suggests a possible indeterminate form \( 0 \cdot \infty \).

Step 2: Convert to a L'Hôpital's Rule Appropriate Form

Re-express the function as a quotient suitable for L'Hôpital's Rule. Rewrite \( x(\ln x)^{2} \) as \( \frac{(\ln x)^{2}}{1/x} \). Now, as \( x \rightarrow 0^{+} \), \( (\ln x)^{2} \) goes to \( +\infty \) and \( 1/x \) also goes to \( +\infty \), creating an \( \infty / \infty \) form suitable for L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to the limit expression: \[\lim _{x \rightarrow 0^{+}} \frac{(\ln x)^{2}}{1/x} = \lim _{x \rightarrow 0^{+}} \frac{2\ln x \cdot \frac{d}{dx}(\ln x)}{-1/x^2}\]Calculate the derivatives:- The derivative of \((\ln x)^{2}\) is \(2\ln x \cdot 1/x = 2\frac{\ln x}{x}\).- The derivative of \(1/x\) is \(-1/x^2\).So we have:\[\lim _{x \rightarrow 0^{+}} \frac{2\frac{\ln x}{x}}{-1/x^2} = \lim _{x \rightarrow 0^{+}} -2x (\ln x)\]

Step 4: Evaluate the New Limit

As \( x \rightarrow 0^{+} \), \( \ln x \to -\infty \), hence \( x \ln x \to 0 \times (-\infty) \) remains an indeterminate form, but shifting the focus to \( -2x(\ln x) \), realize that:\[-2x\ln x \sim -2\cdot 0\cdot (-\infty) \to 0\]Thus, approach again with derivatives if necessary until achieving convergence, which it does towards 0.

Key Concepts

Limits

Limits are foundational in calculus, helping us understand the behavior of functions as they approach a specific point. In this exercise, we are examining the limit of the function \( x(\ln x)^{2} \) as \( x \to 0^{+} \). This notation \( x \rightarrow 0^{+} \) means that \( x \) is approaching zero from the positive side. The limit is a crucial tool for revealing the trend of a function's output as input values get closer and closer to a particular point.

To solve limits effectively, unpacking the components of the expression is helpful. This allows us to predict how changes in \( x \) influence the overall function. For \( x(\ln x)^{2} \), two major components exist: the linear part \( x \), which heads towards zero, and the logarithmic part \((\ln x)^{2} \), which grows towards \( +\infty \) as \( x \to 0^{+} \). Recognizing this demonstrates how limits deal with approaching, not necessarily reaching a definite value.

Indeterminate Forms

Indeterminate forms like \( 0 \cdot \infty \) frequently surface in calculus, indicating an undefined state needing further resolution. Originally, the function \( x(\ln x)^{2} \) as \( x \rightarrow 0^{+} \) combines a term nearing zero with one extending towards infinity. This creates the indeterminate form \( 0 \cdot \infty \), obstructing straightforward limit computation.

To address indeterminate forms, rewriting a product as a quotient can simplify matters. In our scenario, by expressing \( x(\ln x)^{2} \) as \( \frac{(\ln x)^{2}}{1/x} \), we transition to a \( \frac{\infty}{\infty} \) form. This configuration is suitable for applying L'Hôpital's Rule. Recognizing indeterminate forms is essential, as it guides which mathematical tools and transformations to use during problem-solving.

Logarithmic Functions

Logarithmic functions are another key player in this exercise. The natural logarithm \( \ln x \) is defined for positive values of \( x \) and is necessary when dealing with exponential growth. As \( x \) approaches zero from the positive side, \( \ln x \) decreases towards \(-\infty\). This powerful property makes logarithmic functions invaluable for understanding diverse behaviors within a function.

Given the expression \((\ln x)^{2}\), when \( x \) is tiny, the square of a large negative number (like \(-\infty\)) becomes positive, increasing without bound. The understanding of logarithmic functions, especially as they interact in complex expressions, aids in predicting outcomes or interpreting limits. Knowing how to manipulate and apply logarithmic identities can greatly enhance problem-solving efficiency.

Derivatives

Derivatives describe the rate of change of a function with respect to a variable. Applying L'Hôpital's Rule heavily relies on differentiation. The rule allows us to evaluate limits of indeterminate forms, like \( \frac{\infty}{\infty} \), by differentiating the numerator and denominator until a solvable limit appears.

In our exercise, the derivatives of \( (\ln x)^{2} \) and \( 1/x \) are calculated as part of the solution. The derivative of \( (\ln x)^{2} \) is \( 2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x} \), while the derivative of \( 1/x \) is \(-1/x^{2}\). Repeating this differentiation process can simplify the expression further, helping us transition from \(-2x(\ln x) \) to focus on the limit evaluation directly. This illustrates the importance of a strong grasp of derivatives, ensuring accurate and efficient problem resolution in calculus.