Problem 33 - Calculus Single Variable

Question

In each of Exercises \(33-40,\) use the logarithm to reduce the given limit to one that can be handled with l'Hôpital's Rule. \(\lim _{x \rightarrow 0^{+}} x^{\sqrt{x}}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Express the limit in exponential form

The given limit is \( \lim_{x \to 0^+} x^{\sqrt{x}} \). This expression can be rewritten using the exponential function: \( x^{\sqrt{x}} = e^{\sqrt{x} \ln x} \). The limit now becomes \( \lim_{x \to 0^+} e^{\sqrt{x} \ln x} \).

Step 2: Focus on the exponent

Since we are dealing with an exponential function, we can concentrate on finding the limit of the exponent: \( \lim_{x \to 0^+} \sqrt{x} \ln x \). If this limit is \( L \), the overall limit is \( e^L \).

Step 3: Use substitution for simplification

Let \( y = \sqrt{x} \) such that \( x = y^2 \). As \( x \to 0^+ \), \( y \to 0^+ \). The limit now becomes \( \lim_{y \to 0^+} y \ln(y^2) = \lim_{y \to 0^+} 2y \ln y \).

Step 4: Apply l'Hôpital's Rule to the new limit

The limit \( \lim_{y \to 0^+} 2y \ln y \) can be rewritten as \( \lim_{y \to 0^+} \frac{\ln y}{1/y} \). This setup allows for the application of l'Hôpital's Rule as both the numerator and denominator approach infinity. Differentiate the numerator and denominator with respect to \( y \): \( \lim_{y \to 0^+} \frac{\frac{d}{dy}(\ln y)}{\frac{d}{dy}(1/y)} = \lim_{y \to 0^+} \frac{1/y}{-1/y^2} \).

Step 5: Simplify using l'Hôpital's Rule result

Simplifying, we end up with \( \lim_{y \to 0^+} \frac{-y^2}{y} = \lim_{y \to 0^+} -y = 0 \). This suggests that \( L = 0 \).

Step 6: Conclude the original limit

Substituting back, the original limit \( \lim_{x \to 0^+} e^{\sqrt{x} \ln x} \) can be simplified to \( e^0 = 1 \).

Key Concepts

logarithmic limits

Logarithmic limits help us manage complex expressions by using logarithms to simplify the power or exponent. In the given problem, we replaced the original limit involving powers with a limit involving logarithms. This was done by expressing the power in logarithmic form:

• The expression \(x^{\sqrt{x}}\) was rewritten using exponential functions: \(x^{\sqrt{x}} = e^{\sqrt{x} \ln x}\).
• This clever transformation enables us to shift our focus from studying the entire expression to just the exponent: \(\sqrt{x} \ln x\).
• Solving the limit for this exponent can then apply standard calculus methods like l'Hôpital's Rule.

This technique is beneficial when dealing with more complex powers or roots, as it allows us to handle them like simpler expressions purely involving multiplication or division.

exponential functions

Exponential functions are a class of functions where a constant base is raised to a variable exponent. These functions appear frequently in problems involving limits, growth, and decay. Here, we expressed the original power \(x^{\sqrt{x}}\) in terms of the exponential function, transitioning it into the form \(e^{\sqrt{x} \ln x}\). The key properties include:

• Exponential functions have the base \(e\), which is approximately 2.718, known as Euler's number. It is fundamental in continuous growth and decay processes.
• Transformation of the limit problem using the exponential function simplified the analysis by focusing merely on the exponent \(\sqrt{x} \ln x\).
• This approach allows for strategic use of derivative techniques like l'Hôpital's Rule to resolve limits effectively.

Recognizing when to convert expressions using exponential functions is crucial in tackling complex calculus limit problems.

calculus limits

Limits in calculus determine the value that a function approaches as the input approaches some point. In our problem, we needed to find the limit of \(x^{\sqrt{x}}\) as \(x\) approaches 0 from the positive side. Let's break down how limits work here:

• First, we analyzed \(x^{\sqrt{x}}\), recognizing the power configuration to apply logarithmic limits and exponential functions.
• The essence of the problem was to transform it into a simpler form \(e^{\sqrt{x} \ln x}\), ultimately studying the behavior of \(\sqrt{x} \ln x\) as \(x\) approaches 0.
• Substitution and l'Hôpital's Rule were applied to simplify and find the rigorous limit of the difficult expression \(2y \ln y\) by setting \(y = \sqrt{x}\).
• Finally, we translated these approaches back into the original problem by computing \(e^{0} = 1\), the result of the simplified limit.

Understanding these techniques helps solve complex limit problems and is a cornerstone in calculus, critical for students to master when dealing with composite functions and their behaviors at boundary values.