Question

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

Short answer

The limit is 1.

Step-by-step solution

Rewrite Expression for Easier Analysis

To analyze the limit \( \lim_{x \to 0^+} x^x \), we start by rewriting \( x^x \). We know that \( x^x \) can be expressed as an exponential function: \( x^x = e^{x \ln(x)} \). Therefore, our goal is to find \( \lim_{x \to 0^+} e^{x \ln(x)} \).

Simplify the Exponent

Focus on the exponent \( x \ln(x) \). To find its limit as \( x \rightarrow 0^+ \), we rewrite it as \( \frac{\ln(x)}{1/x} \). This transforms the problem into a classic \( \frac{\infty}{\infty} \) form, making L'Hôpital's Rule applicable.

Apply L'Hôpital's Rule

We apply L'Hôpital's Rule to \( \lim_{x \to 0^+} \frac{\ln(x)}{1/x} \). First, we differentiate the numerator and denominator separately: the derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( 1/x \) is \( -\frac{1}{x^2} \). This gives us the new limit \( \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x \).

Evaluate the Simplified Limit

Now, \( \lim_{x \to 0^+} -x = 0 \) as \( x \rightarrow 0^+ \). Thus, the exponent \( \lim_{x \to 0^+} x \ln(x) = 0 \).

Determine the Final Limit

Since \( x^x = e^{x \ln(x)} \) and the exponent \( x \ln(x) \rightarrow 0 \) as \( x \rightarrow 0^+ \), we conclude that \( \lim_{x \to 0^+} e^{x \ln(x)} = e^0 = 1 \). Therefore, \( \lim_{x \to 0^+} x^x = 1 \).

Key concepts

Limits

Limits are a fundamental concept in calculus, describing the value a function approaches as the input approaches a particular point. Calculating limits is crucial to understanding the behavior of functions at points where they might not be explicitly defined. For instance, when analyzing \[ \lim_{x \to 0^+} x^x \], we study how the expression behaves as \(x\) approaches zero from the positive side. The process of finding limits often involves:

• Rewriting expressions for easier analysis.
• Using techniques like substitution or simplification.
• Applying rules such as L'Hôpital's Rule when direct substitution results in indeterminate forms.

In this exercise, rewriting \(x^x\) as \(e^{x \ln(x)}\) allows us to transform the expression to facilitate finding the limit. Recognizing and handling indeterminate forms, like \(\frac{\infty}{\infty}\), is key when working with limits.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for solving indeterminate forms involving limits, such as \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). When a limit calculation leads to such forms, the rule helps by allowing us to differentiate the numerator and the denominator separately. In this context, the rule assists in managing the expression \(x \ln(x)\), rewritten as \(\frac{\ln(x)}{1/x}\). Using L'Hôpital's Rule, we differentiate both parts:

• For \(\ln{x}\), the derivative is \(\frac{1}{x}\).
• For \(\frac{1}{x}\), the derivative is \(-\frac{1}{x^2}\).

Post differentiation, the new limit \(\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x\) simplifies the problem significantly. Recognizing when and how to apply L'Hôpital's Rule is fundamental, simplifying complex limit calculations and leading to those crucial solutions.

Exponential Function

Exponential functions play a critical role in many areas of mathematics, characterized by the constant "e", approximately equal to 2.71828. In this exercise, the goal was to assess the limit involving the exponential expression \(e^{x \ln(x)}\) as \(x\) approaches zero from the positive side. Understanding that the exponential function can model growth rates and decay is essential. The expression \(x^x\) can be complex, but rewriting it as \(e^{x \ln(x)}\) by leveraging the properties of \(e\) simplifies analysis and reveals more about the underlying function. By recognizing that \(e^0 = 1\), it becomes clear why the limit of \(x^x\) as \(x\) approaches 0 from the positive side is 1 once the exponent goes to zero. Exponential functions are not only mathematically intriguing but also have real-world applications, from calculating compound interest to modeling populations or radioactive decay. A deeper grasp of how exponential functions operate aids in comprehending both theoretical and practical problems.