Problem 64 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow 0^{+}} x(\ln x)^{2}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Expression

We need to find the limit of the given expression \[\lim_{x \to 0^+} x(\ln x)^2.\]As \(x\) approaches 0 from the positive side, \(\ln x\) approaches \(-\infty\). Hence, \((\ln x)^2\) approaches \(\infty\). Meanwhile, \(x\) approaches 0, which suggests this is an indeterminate form of the type \(0 \cdot \infty\).

Step 2: Transform the Expression

We need to transform the indeterminate form into a limit that can be evaluated. We rewrite the expression as\[\lim_{x \to 0^+} \frac{(\ln x)^2}{\frac{1}{x}}.\]Now we have a \(\frac{\infty}{\infty}\) form, which is suitable for using L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to the transformed expression. This rule states that \[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.\]Differentiate the numerator and the denominator:- Numerator: the derivative of \((\ln x)^2\) is \(2 \cdot \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}\).- Denominator: the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\).Substitute these into the limit:\[\lim_{x \to 0^+} \frac{\frac{2 \ln x}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -2x \ln x.\]This again simplifies to an indeterminate form \(0 \cdot \infty\).

Step 4: Reapply L'Hôpital's Rule

Notice that with the new form \(-2x \ln x\), we have another \(0 \cdot \infty\) situation. Rewrite this limit as:\[\lim_{x \to 0^+} \frac{-2 \ln x}{\frac{1}{x}}.\]The expression is again of the form \(\frac{\infty}{\infty}\). Apply L'Hôpital's Rule one more time.- Numerator: the derivative of \( -2 \ln x \) is \( \frac{-2}{x} \).- Denominator: the derivative of \( \frac{1}{x} \) is \(\frac{-1}{x^2}\).This yields:\[\lim_{x \to 0^+} \frac{\frac{-2}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} 2x = 0.\]

Step 5: Conclusion

After applying L'Hôpital's Rule twice, we have simplified the original indeterminate form to a straightforward limit. The limit evaluates to 0 as \(x\) approaches 0 from the positive side.

Key Concepts

L'Hôpital's Rule

When faced with limits that result in undefined expressions, often called indeterminate forms, L'Hôpital's Rule becomes a highly useful tool.

It provides a straightforward method to evaluate certain limits. Essentially, L'Hôpital's Rule states that if we have a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately, then take the limit again.

• It is crucial that the original limit is indeterminate and both functions in the fraction are differentiable.
• The derivatives must again form a valid limit structure; if not, you might have to resort to other methods or repeat the rule if the same forms appear.

In our example, we used L'Hôpital's Rule to resolve a \( \frac{\infty}{\infty} \) form that emerged after transforming \( 0 \cdot \infty \) into a rational form. By systematically differentiating and simplifying, we arrived at a solvable limit. Remember, differentiating correctly and keeping track of each component is vital for successfully applying L'Hôpital's Rule.

Indeterminate Forms

In calculus, indeterminate forms are expressions encountered when attempting to calculate limits that do not have a clear, defined answer at first glance. They require additional techniques to resolve.

The most common indeterminate forms you'll encounter include:

• \( 0/0 \)
• \( \infty/\infty \)
• \( 0 \cdot \infty \)
• \( 0^0 \)
• \( \infty - \infty \)
• \( 1^\infty \)
• \( \infty^0 \)

These forms indicate that a straightforward approach won't work, and alternative strategies like algebraic manipulation or L'Hôpital's Rule are necessary.

In the exercise provided, initially, \( x(\ln x)^2 \) became indeterminate as \( x \to 0^{+} \), showing a \( 0 \cdot \infty \) form. By transforming it into \( \frac{(\ln x)^2}{\frac{1}{x}} \), we ended up with a \( \frac{\infty}{\infty} \) form, clearly suitable for L'Hôpital's Rule. Understanding these forms helps in quickly identifying the appropriate methods to simplify and solve a limit problem.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithmic function with the base \( e \), where \( e \approx 2.71828 \). It is one of the most important functions in calculus due to its unique properties.

Important traits of the natural logarithm include:

• **Growth behavior:** As \( x \to 0^{+} \), \( \ln x \to -\infty \), indicating rapid negative growth.
• **Continuity and differentiability:** \( \ln x \) is continuous and differentiable wherever it is defined, specifically on \( (0, \infty) \).
• **Derivative:** The derivative of \( \ln x \) is \( \frac{1}{x} \).

In our problem, \( \ln x \) was squared, intensifying the negative aspect as \( x \) approaches zero. Its behavior significantly influenced the limit, particularly causing the expression to shift through different indeterminate forms.

Understanding how natural logarithms operate aids in grasping why certain transformations and solutions are necessary, particularly when dealing with limits involving logarithmic expressions.