Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 0}|x|^{x} $$

Step-by-Step Solution

Answer

1

Step 1: Understand the Expression

We are asked to find the limit of the expression \(|x|^x\) as \(x\) approaches 0. The expression involves an absolute value and an exponent where \(x\) is both the base and the exponent.

Step 2: Analyze Base and Exponent as x approaches 0

As \(x\) approaches 0, \(|x|\) approaches 0. However, \(x\), the exponent, also approaches 0. Thus, we have an indeterminate form of type \(0^0\), which requires further analysis.

Step 3: Transform Expression for Analysis

Consider the logarithm of the expression: \(x \cdot \ln|x|\). The limit of the original expression \(|x|^x\) can thus be found by investigating \(e^{x \ln|x|}\). This helps simplify handling the indeterminate form.

Step 4: Evaluate the Inner Limit

Evaluate the limit \(L = \lim_{x \rightarrow 0} x \ln|x|\). We rewrite \(x \ln|x|\) as \(\frac{\ln|x|}{1/x}\), which is now a \(\frac{-\infty}{\infty}\) form.

Step 5: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to \(\lim_{x \rightarrow 0} \frac{\ln|x|}{1/x}\). Derivative of numerator \(\ln|x|\) is \(1/x\) (consider positive and negative \(x\)), and the derivative of denominator \(1/x\) is \(-1/x^2\).

Step 6: Simplify After Applying L'Hôpital's Rule

Simplifying, we have \(\lim_{x \rightarrow 0} \frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0} -x = 0\). Therefore, \(L = 0\).

Step 7: Compute the Limit of the Original Expression

Since \(\lim_{x \rightarrow 0} x \ln|x| = 0\), we conclude that \(\lim_{x \rightarrow 0} e^{x \ln|x|} = e^0 = 1\). Thus, \(\lim_{x \rightarrow 0}|x|^x = 1\).

Key Concepts

Indeterminate Forms

When dealing with calculus limits, an indeterminate form represents expressions where the limits cannot be determined directly by substitution. These forms arise when specific algebraic or trigonometric expressions approach a certain value that makes the limit appear undefined or ambiguous.

Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( \infty - \infty \)
• \( 0 \times \infty \)
• \( 0^0 \)
• \( \infty^0 \)
• \( 1^\infty \)

Each of these forms requires further manipulation or application of specific rules in order to evaluate the limit. The exercise touched upon the indeterminate form \(0^0\), prompting a more intricate approach beyond simple substitution, such as transformations or the application of other mathematical techniques.

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool in calculus for evaluating limits that initially present as indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The basic premise is to differentiate the numerator and the denominator separately and then re-evaluate the limit.

Here’s the basic application of L'Hôpital's Rule:

• Identify the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Differentiate the numerator and the denominator.
• Reevaluate the limit using these new derivatives.
• Repeat the process if necessary.

The exercise utilized this rule by transforming the expression \(x \ln|x|\) into \(\frac{\ln|x|}{1/x}\), and applying L'Hôpital's Rule to evaluate the limit, resulting in a solvable expression leading to the final result.

Limit Evaluation

The process of limit evaluation involves determining the value that a function approaches as the input approaches a certain point. This calculation is central to the study of calculus since it helps understand the behavior of functions at specific points or as they trend towards infinity.

When approaching a limit problem, consider:

• Substitution: First step is always a straightforward substitution to see if the function yields a clear value.
• Simplification: Manipulate the function using algebraic principles to simplify it, removing any complexities or indeterminate forms.
• Application of Rules: Use rules like L'Hôpital's Rule or logarithmic transformations to solve complex limits.
• Re-Expression: Use logarithms or other methods to transform a limit that seems unsolvable at first glance.

In the exercise, limit evaluation was approached by substituting and simplifying the problem using logarithmic transformations and the subsequent application of L'Hôpital's Rule to arrive at \(\lim_{x \rightarrow 0}|x|^x = 1\).

Absolute Value in Limits

Absolute value often appears in calculus problems involving limits, adding an additional layer of consideration when evaluating a limit. Absolute value affects how the function behaves around specific points, particularly when considering the behavior from the left-hand and right-hand collisions with the limit point.

Important aspects of absolute value in limits include:

• Considering the two-sided behavior as \(x\) approaches the limit point, differentiating the analysis for positive and negative sides if necessary.
• Using the definition of absolute value, \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\), to reframe the problem if derivatives need to be taken.
• Understanding that absolute value affects the limits by potentially creating piecewise functions.

In the exercise, \(|x|^x\), the absolute value ensures that the base of the exponential function is always positive, thus simplifying the treatment of the function consistent across all values but necessitates precise handling during derivative evaluation in L'Hôpital's rule.