Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0^{+}}\left(x^{1 / 2} \ln x\right)$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form of the Expression

We need to determine the form of the expression as \( x \) approaches \( 0^{+} \). Substituting \( x = 0^{+} \) in \( x^{1/2} \ln x \), we see that \( x^{1/2} \rightarrow 0 \) and \( \ln x \rightarrow -\infty \). Therefore, the expression takes the form \( 0 \times (-\infty) \), which is an indeterminate form.

Step 2: Rewrite to Apply l'Hôpital's Rule

We need to convert the expression into a quotient to apply l'Hôpital's Rule. Rewrite \( x^{1/2} \ln x \) as \( \frac{\ln x}{1/x^{1/2}} \). This is now in the \( \frac{-\infty}{\infty} \) form, an indeterminate form suitable for l'Hôpital's Rule.

Step 3: Find Derivatives of Numerator and Denominator

Find the derivative of the numerator \( \ln x \), which is \( \frac{1}{x} \), and the derivative of the denominator \( 1/x^{1/2} \), which is \( -\frac{1}{2} x^{-3/2} \).

Step 4: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule to the expression:\[\lim _{x \rightarrow 0^{+}} \frac{\ln x}{1/x^{1/2}} = \lim _{x \rightarrow 0^{+}} \frac{1/x}{-\frac{1}{2} x^{-3/2}} = \lim _{x \rightarrow 0^{+}} \frac{-2x^{3/2}}{x}\]Simplify the expression to obtain:\[\lim _{x \rightarrow 0^{+}} -2x^{1/2}\]

Step 5: Evaluate the Simplified Limit

Substitute \( x = 0^{+} \) into \( -2x^{1/2} \). Since \( x^{1/2} \rightarrow 0 \) as \( x \rightarrow 0^{+} \), the whole expression \( -2x^{1/2} \rightarrow 0 \). Hence, the limit is \( 0 \).

Key Concepts

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating a limit doesn't immediately lead to a specific value. These forms signal ambiguity, meaning that more analysis is needed to find the limit. When substituting into a limit results in forms like 0/0, ∞/∞, 0·∞, ∞−∞, or similar, these are considered indeterminate.
In our exercise, as we substituted into our expression as \( x \) approaches \( 0^+ \), we encountered \( 0 \times (-\infty) \). This is an indeterminate form because multiplying zero by negative infinity doesn't yield a clear numerical value.
Recognizing such forms is crucial before applying rules like l'Hôpital's because it indicates the need for transformation or further steps to properly evaluate the limit.

Limits in Calculus

Limits in calculus are foundational. They help us understand the behavior of functions as the input approaches a certain value. Essentially, limits are used to define both the continuity and differentiability of functions.
In this exercise, we needed to determine the behavior of \( x^{1/2} \ln x \) as \( x \rightarrow 0^+ \). Direct substitution didn’t work due to the indeterminate form. Thus, our task was to manipulate the expression into a form where we could find the limit.Finding limits often involves algebraic manipulation, taking advantage of limit properties, or applying rules such as l'Hôpital's when faced with particular indeterminate forms. In our example, rewriting the product \( x^{1/2} \ln x \) into a quotient allowed us to find the limit using l'Hôpital's Rule which is specifically designed to handle indeterminate forms like \( \frac{-\infty}{\infty} \).

Derivatives

Derivatives, one of the core concepts in calculus, represent the rate of change of a function. They provide tools for analyzing how functions behave and can be used to solve various practical problems.
In this problem, derivatives were crucial for applying l'Hôpital's Rule. After rewriting \( x^{1/2} \ln x \) into a quotient form, we needed the derivatives of the numerator and the denominator. The derivative of \( \ln x \) is \( \frac{1}{x} \), and for \( 1/x^{1/2} \), the derivative is \( -\frac{1}{2} x^{-3/2} \).Calculating these derivatives allowed us to form a new limit expression that L'Hôpital's rule could be applied to. The limits of these derivatives helped simplify the expression and ultimately find the solution.