Question

Determine whether you can apply L'Hôpital's rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L'Hôpital's rule. $$ \lim _{x \rightarrow 0^{+}} x^{2} \ln x $$

Short answer

L'Hôpital's Rule cannot be directly applied but can be used after rewriting the limit as \( \frac{\ln x}{1/x^2} \) to get 0.

Step-by-step solution

Analyze the expression type

The expression given is \( x^{2} \ln x \) as \( x \to 0^+ \). Let's check what happens to each term: \( \ln x \to -\infty \) as \( x \to 0^+ \), and \( x^2 \to 0 \). This results in an indeterminate form of type \( (0)(-\infty) \).

Determine if L'Hôpital's Rule is applicable

L'Hôpital's Rule is applicable for limits that result in forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The current form, \( (0)(-\infty) \), does not directly satisfy these conditions, so we cannot apply L'Hôpital's Rule directly.

Alter the expression to apply L'Hôpital's Rule

To use L'Hôpital's Rule, rewrite the expression so it becomes a quotient. Rewrite \( x^2 \ln x \) as \( \frac{\ln x}{1/x^2} \). As \( x \to 0^+ \), this becomes \( \frac{-\infty}{\infty} \), a form suitable for L'Hôpital's Rule.

Apply L'Hôpital's Rule

Compute the derivatives of the numerator and denominator. The derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of \( \frac{1}{x^2} \) is \( -\frac{2}{x^3} \). Now the limit is rewritten as: \[ \lim_{x \to 0^+} \frac{1/x}{-2/x^3} = \lim_{x \to 0^+} \frac{-x^2}{2} = 0. \]

Verify the calculation

The evaluation shows that the limit exists and is 0. This confirms that with the altered form \( \frac{\ln x}{1/x^2} \), L'Hôpital's Rule was applied correctly, leading to the solution.

Key concepts

Indeterminate Forms

When evaluating limits in calculus, you may encounter expressions that don't readily simplify to a single answer. These are termed "indeterminate forms." Common examples include expressions such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others. Although these forms seem paradoxical, there are mathematical strategies to resolve them to find a definite limit value.

For example, in the exercise we have the expression \( x^2 \ln x \) as \( x \to 0^+ \). As \( x \) approaches zero, \( x^2 \) tends to 0 and \( \ln x \) trends to \(-\infty\), resulting in the indeterminate form \((0)(-\infty)\). While it seems difficult to predict the behavior of such a product, calculus techniques, particularly L'Hôpital's Rule, help us navigate these tricky limits.

Recognizing these forms early allows you to adjust or manipulate expressions to make them suitable for calculus tools, making seemingly impossible evaluations possible.

Limit Evaluation

In calculus, evaluating a limit means determining the value that a function approaches as the input approaches some point. It's a fundamental concept that helps us understand the behavior of functions at points of interest, such as where they may not be defined or as they approach infinity.

Not every limit is straightforward to solve, like the exercise where the limit \( \lim _{x \rightarrow 0^{+}} x^{2} \ln x \) needs evaluation. Initially, direct substitution gives us an indeterminate form, requiring an alternate approach.

One common method is to rewrite the challenging expression in a form that makes applying calculus techniques—like L'Hôpital's Rule—possible. In our case, rewriting \( x^2 \ln x \) as \( \frac{\ln x}{1/x^2} \) helps transition the troublesome product into a quotient that simplifies the process of limit evaluation.

Calculus Techniques

L'Hôpital's Rule is a powerful tool for solving limits that result in indeterminate forms of the type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule provides a straightforward procedure: take the derivative of the numerator and the derivative of the denominator, then evaluate the limit again.

However, the rule isn't directly applicable to products such as \((0)(-\infty)\). Instead, the expression must first be reconfigured into a quotient. This often requires clever algebraic manipulation, such as expressing \( x^2 \ln x \) as \( \frac{\ln x}{1/x^2} \). Once in this form, as noted in the exercise, it becomes a \( \frac{-\infty}{\infty} \) scenario, thus allowing L'Hôpital's Rule to be applied.

After derivatives are computed—a straightforward procedure where the derivative of \( \ln x \) is \( \frac{1}{x} \) and for \( 1/x^2 \) is \( -\frac{2}{x^3} \)—the revised limit can be accurately evaluated. Here, the problem resolves to \( \lim_{x \to 0^+} \frac{-x^2}{2} \), confirming the limit is \( 0 \).

• Recognizing when to use L'Hôpital's Rule is crucial for transforming problems into more manageable forms.
• Always ensure derivatives are computed correctly to maintain the integrity of the solution.
• Check your final result to verify accuracy and consistency with mathematical principles.