Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0^{+}} x^{2 x} $$

Short answer

The limit is 1.

Step-by-step solution

Rewrite the Expression

To evaluate the limit of \( \lim_{x \to 0^+} x^{2x} \), it's useful to rewrite the exponential expression using natural logarithms for easier manipulation: \( x^{2x} = e^{2x \ln x} \).

Evaluate the Exponent

We focus on the exponent \( 2x \ln x \). To do this, we will evaluate the limit of this expression as \( x \to 0^+ \). Express the limit as \( \lim_{x \to 0^+} 2x \ln x \).

Apply L'Hôpital's Rule

Recognize that \( 2x \ln x \) can be rewritten as \( \frac{2 \ln x}{1/x} \), which is of the indeterminate form \( \frac{-\infty}{\infty} \); hence, L'Hôpital's Rule applies. Differentiate the numerator and the denominator: \( \lim_{x \to 0^+} \frac{2 \cdot \frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -2x \).

Evaluate the New Limit Expression

Now evaluate \( \lim_{x \to 0^+} -2x \). As \( x \to 0^+ \), \( -2x \to 0 \). So, \( \lim_{x \to 0^+} 2x \ln x = 0 \).

Substitute Back to Original Expression

Substitute back the evaluated limit into the original expression: Since \( 2x \ln x \to 0 \), then \( e^{2x \ln x} \to e^0 = 1 \). Therefore, \( \lim_{x \to 0^+} x^{2x} = 1 \).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits that result in indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule allows us to differentiate the numerator and denominator of a function separately to evaluate such tricky limits.
When you encounter one of these indeterminate forms, apply L'Hôpital's Rule by taking the derivative of the top and bottom functions. Continue applying the rule until the expression can be evaluated directly.
Here’s how it works:

• Identify if the limit is in an indeterminate form.
• Differentiation: Compute the derivative of the numerator and the derivative of the denominator.
• Re-evaluate the limit with these derivatives.

It's essential to remember that L'Hôpital's Rule only works if differentiation leads to a simplification that allows for limit evaluation. In the context of our original exercise, we faced the indeterminate form \(-\infty/\infty\), making L'Hôpital's Rule an ideal choice to handle the situation.

Exponential Functions

Exponential functions have the form \(b^{x}\) where \(b\) is a positive real number, and \(x\) is the exponent. These functions grow rapidly, and their behavior is crucial in calculus. In this exercise, our expression \(x^{2x}\) was initially difficult to evaluate, reflecting the complexity of dealing with variable bases and exponents directly.
To simplify the evaluation of limits for exponential functions, we often turn to another mathematical tool: the natural logarithm. By rewriting the exponential expression \(x^{2x}\) using the identity \(x^{2x} = e^{2x \ln x}\), we can work more easily in terms of linear expressions. The natural logarithm simplifies exponents, providing a straightforward means to resolve limits.
Understanding exponential functions is key not just for solving limits problems but also for modeling real-world situations like population growth or radioactive decay.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is the inverse function of the exponential function \(e^x\). Understanding this concept is pivotal in calculus, particularly when dealing with transformation and evaluation of exponential expressions.
In the problem we addressed, the natural logarithm enabled us to rewrite \(x^{2x}\) into an expression involving \(e\), a mathematically universal constant approximately equal to 2.71828. This transformation simplifies complexity, turning a power function into a more manageable form.
The steps are as follows:

• Recognize the necessity of simplification with \(\ln x\) for complex expressions.
• Convert power expressions using the rule \(x^{a} = e^{a \ln x}\), which allows us to tap into the properties of the exponential function.
• In the rewritten \(e^{2x \ln x}\), the behavior as \(x \to 0^{+}\) could be evaluated directly using known properties of continuous and differentiable functions.

The natural logarithm not only helps in solving problems involving limits but proves indispensable in various calculus operations, especially in integration and differentiation.