Problem 41 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}} x^{2 x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the expression using logarithms

We have the expression \(x^{2x}\). We start by taking the natural logarithm of the limit to simplify our calculations: \( \ln L = \lim_{x \to 0^+} \ln (x^{2x}) \).We can rewrite this as \( \ln L = \lim_{x \to 0^+} 2x \cdot \ln x \).

Step 2: Apply L'Hospital's Rule

The expression \( \lim_{x \to 0^+} 2x \ln x \) yields the indeterminate form \( 0 \cdot -\infty \). Rewrite it as a fraction to apply L'Hospital's rule.\( \lim_{x \to 0^+} \frac{\ln x}{1/(2x)} \). This is in the form \( \frac{-\infty}{\infty} \).

Step 3: Differentiate the Numerator and the Denominator

Differentiate the numerator and the denominator:The derivative of \( \ln x \) is \( \frac{1}{x} \). The derivative of \( \frac{1}{2x} \) is \( -\frac{1}{2x^2} \).

Step 4: Apply L'Hospital's Rule

Apply L'Hospital's Rule:\[ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2x^2}} \]which simplifies to:\[ \lim_{x \to 0^+} \frac{-2x}{1} = \lim_{x \to 0^+} -2x \].

Step 5: Evaluate the Limit after Differentiation

Since \( -2x \to 0 \) as \( x \to 0^{+} \), the limit is 0.Therefore, \( \ln L = 0 \).

Step 6: Find the Exponential to Get Back to L

Given \( \ln L = 0 \), it follows that \( L = e^0 = 1 \).

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions in calculus that arise when assessing certain limits which do not initially provide sufficient information to determine a clear outcome. These forms often appear as ratios, products, or powers that behave unpredictably at specific points, such as when approaching zero or infinity. In our exercise, as we evaluate the limit \( \lim _{x \rightarrow 0^{+}} x^{2x} \), the expression converts to \( x^{2x} \), yielding an indeterminate form. Indeterminate forms typically include 0/0, \( \infty/\infty \), 0 \( \times \infty \), 1\(^\infty\), 0\(^0\), \( \infty \)^0, and \( \infty - \infty \).
This specific problem results in an indeterminate form of \( 0 \cdot -\infty \) when rewritten as \( 2x \cdot \ln x \). Solving these forms requires specialized strategies, such as rewriting the expression and using l'Hospital's Rule.
This rule aids in evaluating limits where standard simplification techniques fall short, providing a structured way forward when traditional algebraic manipulation does not suffice.

Differentiation

Differentiation is a fundamental mathematical tool used to determine the rate at which a function changes at any given point. It involves computing the derivative, which is essentially the slope of the tangent line to the graph of the function. Derivatives are crucial for applying l'Hospital’s Rule, as they allow us to simplify limits involving indeterminate forms. In the solution provided, numerical differentiation of both the top and bottom of a fraction helps recast the problem into a solvable form.
When finding the limit, we differentiate the numerator and the denominator:

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• The derivative of \( \frac{1}{2x} \) is \( -\frac{1}{2x^2} \).

By differentiating these expressions and applying l'Hospital's Rule, the complex limit simplifies significantly. The rules of differentiation, especially as they pertain to natural logarithms and rational functions, are central to resolving bounds that appear tricky at face value. This exercise demonstrates the power of differentiation to transform complex expressions into easily manageable ones.

Logarithmic Differentiation

Logarithmic differentiation is a technique often used when differentiating complex expressions, particularly when dealing with variables raised to the power of variables, such as \( x^{2x} \). This method relies on taking the natural logarithm of both sides of an equation to simplify the differentiation process, turning products and powers into sums and simpler expressions, respectively.
In our exercise, \( x^{2x} \) is transformed by taking its natural logarithm to become \( 2x \cdot \ln x \). This step leverages the properties of logarithms to convert an expression initially appearing as an indeterminate power into a product that's simpler to analyze.
The approach effectively breaks down a complex function into more manageable parts, allowing us to apply differentiation methods effectively. Logarithmic differentiation simplifies the expression significantly, enabling a straightforward path to solving the indeterminate form with the application of l'Hospital's Rule.