Problem 63 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Limit.

We need to evaluate the limit \( \lim_{x \to 0^{+}} x^2 \ln x \). This function involves a product of \( x^2 \) and \( \ln x \), where \( x \) approaches zero from the right.

Step 2: Behavior of Each Component

As \( x \to 0^{+} \), \( x^2 \to 0^{+} \) and \( \ln x \to -\infty \). We determined that \( x^2 \) approaches zero and \( \ln x \) goes to negative infinity.

Step 3: Indeterminate Form

The limit is in the indeterminate form 0 \( \times \) \(-\infty\), so we should modify it. Rewrite \( x^2 \ln x \) as \(\frac{\ln x}{1/x^2} \) to apply L'Hôpital's Rule.

Step 4: Apply L'Hôpital's Rule

Reformulate the limit so that it appears as \( \frac{-\infty}{\infty} \), which is suitable for L'Hôpital's Rule. The limit becomes \( \lim_{x \to 0^{+}} \frac{\ln x}{1/x^2} \).

Step 5: Differentiate Numerator and Denominator

Differentiate the numerator (\( \ln x \)) to obtain \( \frac{1}{x} \), and the denominator (\( 1/x^2 \)) to obtain \( -\frac{2}{x^3} \).

Step 6: Evaluate the Modified Limit

The limit becomes \( \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \to 0^{+}} -\frac{x^2}{2} \), which approaches 0 as \( x \to 0^{+} \).

Step 7: Conclusion

Since the function simplifies to \( -\frac{x^2}{2} \), as \( x \to 0^{+} \), the limit evaluates to 0.

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions where direct substitution in limits leads to uncertainty, often showing as expressions like \( 0 \cdot \infty \), \( \frac{0}{0} \), or \( \frac{\infty}{\infty} \). These forms occur naturally in calculus when evaluating limits and require special techniques to resolve.
For example, in the original exercise, as \( x \) approaches zero from the right, \( x^2 \) tends to zero and \( \ln x \) approaches negative infinity, creating an indeterminate form \( 0 \times -\infty \). This means that it’s not immediately clear what the limit of their product is, necessitating some manipulation, such as rewriting products into quotient form, to make it solvable using other methods like L'Hôpital's Rule.
Understanding these forms is crucial because it prevents incorrect assumptions about the behavior and outcome of limits, especially as they involve functions that may seem to produce straightforward results but actually possess hidden complexities.

L'Hôpital's Rule

L'Hôpital's Rule provides a powerful tool for resolving indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When we encounter these forms while evaluating limits, L'Hôpital’s Rule suggests differentiating the numerator and denominator separately and then taking the limit again.
For example, in evaluating the limit \( \lim_{x \to 0^{+}} \frac{\ln x}{1/x^2} \) which is derived from \( x^2 \ln x \), we apply L'Hôpital's Rule. First, differentiate \( \ln x \) to \( \frac{1}{x} \) and \( 1/x^2 \) to \( -\frac{2}{x^3} \).
After differentiation, the limit becomes \( \lim_{x \to 0^{+}} -\frac{x^2}{2} \). Now the limit resolves easily to 0. This technique simplifies otherwise complex expressions into more manageable forms, enabling easier evaluation.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. It is a fundamental function used extensively in calculus and mathematical modeling.
The function \( \ln x \) has interesting properties, particularly its behavior as \( x \) approaches critical points like zero and infinity:

• As \( x \to 0^{+} \), \( \ln x \to -\infty \). This property is pivotal when dealing with limits that approach zero because it can significantly influence the outcome based on the rate of decay.
• \( \ln 1 = 0 \), which is a useful point of reference when shifting logarithmic graphs.

Understanding the behavior of natural logarithms helps in recognizing and solving problems involving exponential growth and decay, including those that result in indeterminate limit forms.

Limit Evaluation

Limit evaluation involves determining the value that a function "approaches" as the input approaches a specific point. This is crucial in understanding the behavior of functions in calculus, especially when direct substitution does not work due to indeterminate forms.
Evaluating limits requires techniques such as:

• Direct Substitution: This works in simple cases where plugging in the value for \( x \) yields a solution.
• Algebraic Manipulation: Adjusting the expression, such as factorizations or simplifications, making it easier to understand the overall behavior.
• Special Methods: Techniques like L'Hôpital's Rule, are often needed when encountering complex indeterminate forms.

Limits are foundational for defining derivatives and integrals, making them integral (pun intended) to calculus. In our case, the use of various techniques transformed an indeterminate form to a simple expression that evaluated the original limit to 0, reflecting the finesse required in calculus to find clear answers from potentially confusing expressions.