Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}}(3 x)^{x^{2}} $$

Short answer

The limit is 1.

Step-by-step solution

Identify the Form of the Limit

First, observe the function and the given limit. We are asked to find \( \lim_{x \to 0^+}(3x)^{x^2} \). As \( x \to 0^+ \), the base \( 3x \to 0^+ \) and the exponent \( x^2 \to 0 \). The expression is of the form \( 0^0 \), which is an indeterminate form.

Set Up l'Hôpital's Rule Conditions

To apply l'Hôpital's Rule, we need to manipulate the limit into a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form. Start by taking the natural logarithm of the expression: let \( y = (3x)^{x^2} \). Then \( \ln(y) = x^2 \ln(3x) \). Thus, we have \( \lim_{x \to 0^+} \ln(y) = \lim_{x \to 0^+} x^2 \cdot \ln(3x) \). This needs to be rewritten to fit a l'Hôpital form.

Rewrite for l'Hôpital's Rule

Re-express the limit \( \lim_{x \to 0^+} x^2 \cdot \ln(3x) \) as a fraction. Notice that as \( x \to 0^+ \), \( \ln(3x) \to -\infty \) and \( x^2 \to 0 \), suggesting an indeterminate form of \( 0 \cdot -\infty \). Rewrite this as: \( \lim_{x \to 0^+} \frac{\ln(3x)}{1/x^2} \). Now we have a \( \frac{-\infty}{\infty} \) form, suitable for l'Hôpital's Rule.

Apply l'Hôpital's Rule

Differentiate the numerator and the denominator separately. The derivative of the numerator, \( \ln(3x) \), is \( \frac{1}{3x} \times 3 = \frac{1}{x} \). The derivative of the denominator, \( 1/x^2 \), is \( -2/x^3 \). Apply l'Hôpital's Rule:\[\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \to 0^+} \frac{-x^2}{2} = 0. \]Thus, \( \lim_{x \to 0^+} \ln(y) = 0 \).

Exponentiate to Find the Limit

Since we had \( y = (3x)^{x^2} \) and \( \ln(y) \) approached 0 as \( x \to 0^+ \), exponentiating both sides gives \( y = e^0 = 1 \). Therefore, \( \lim_{x \to 0^+} (3x)^{x^2} = 1 \).

Key concepts

Indeterminate Forms

Indeterminate forms are expressions in calculus where substitution results in an undefined answer. Common types include forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and \( 0^0 \). In this exercise, the expression \( (3x)^{x^2} \) as \( x \to 0^+ \) appears as the indeterminate form \( 0^0 \). This is because the base \( 3x \) approaches 0 while the exponent \( x^2 \) also trends towards 0. Indeterminate forms are crucial in determining when l'Hôpital's Rule is applicable. Once identified, we look for ways to transform the expression to use l'Hôpital's Rule effectively. Recognizing these forms helps us choose the right mathematical method to find a limit, moving towards clearer, solvable forms.

Limits

Limits are fundamental in calculus, serving to define the behavior of functions as they approach specific points. In this problem, we examine the limit \( \lim_{x \to 0^+} (3x)^{x^2} \). The notation \( x \to 0^+ \) indicates we are considering values of \( x \) approaching 0 from the positive side. This approach helps us explore what happens to our expression under these conditions, and understand functions even when direct substitution doesn't provide answers due to indeterminate forms. Limits are essential in fields such as calculus to comprehend continuity, derivatives, and integrals. They allow us to predict and interpret trends and values where direct calculation may not initially be possible.

Calculus

Calculus is a branch of mathematics that studies how things change. It provides us tools to calculate and understand changes using limits, derivatives, and integrals. In this specific exercise, calculus is used to transform an indeterminate form into a solvable one. We apply l'Hôpital's Rule, which is a derivative-based technique to find limits of fractions that result in indeterminate forms like \( 0/0 \) or \( \infty/\infty \). By differentiating the numerator and the denominator separately, we aim to simplify the problematic form to one we can evaluate directly. This application showcases how calculus can solve complex problems by breaking them into manageable parts using established rules and techniques.

Natural Logarithms

Natural logarithms, denoted \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). They play a pivotal role in calculus for simplifying expressions, especially those involving exponentiation. In our solution, we use natural logarithms to aid in the application of l'Hôpital's Rule. By taking the natural logarithm of both sides of the expression \( y = (3x)^{x^2} \), we transform it into \( \ln(y) = x^2 \ln(3x) \). This transformation helps convert the original indeterminate form \( 0^0 \) into a fraction \( \frac{-\infty}{\infty} \), suitable for l'Hôpital's Rule. Natural logarithms simplify multiplication into addition, which is often easier to analyze mathematically, particularly when finding limits.