Problem 73 - Calculus Single Variable

Question

Use l'Hôpital's Rule to evaluate the one-sided limit. \(\lim _{x \rightarrow 0^{+}} \frac{2 \ln (x)}{\ln (2 x)}\)

Step-by-Step Solution

Answer

The limit evaluates to 4.

Step 1: Verify the Indeterminate Form

To apply l'Hôpital's Rule, we first need to ensure that the given limit is an indeterminate form. As \(x\) approaches \(0^{+}\), both \(\ln(x)\) and \(\ln(2x)\) approach \(-\infty\). Therefore, the fraction \(\frac{2 \ln(x)}{\ln(2x)}\) is in the indeterminate form \(\frac{-\infty}{-\infty}\).

Step 2: Differentiate the Numerator and Denominator

Apply l'Hôpital's Rule by differentiating the numerator and the denominator. The derivative of the numerator, \(2 \ln(x)\), is \(\frac{2}{x}\). The derivative of the denominator, \(\ln(2x)\), is \(\frac{1}{2x}\) because the derivative of \(\ln(u)\) is \(\frac{1}{u}\) times the derivative of \(u\).

Step 3: Simplify the Expression

After differentiation, simplify the fraction:\[\lim_{x \rightarrow 0^{+}} \frac{\frac{2}{x}}{\frac{1}{2x}}\]This simplifies to:\[\lim_{x \rightarrow 0^{+}} \frac{2}{x} \times \frac{2x}{1} = 4\]

Step 4: Evaluate the Limit

Now that we have simplified to a constant term, evaluate the limit. The expression simplifies to \(4\), which is independent of \(x\). Thus:\[\lim_{x \rightarrow 0^{+}} \frac{2 \ln(x)}{\ln(2x)} = 4\]

Key Concepts

Calculus

Calculus is a branch of mathematics concerned with the study of change. It consists of two main parts: differential and integral calculus. These facilitate the understanding of rates of change and the accumulation of quantities.

• Differential Calculus: Focuses on the concept of a derivative, which is a measure of how a function changes as its input changes. It's the mathematical way to express and analyze change.
• Integral Calculus: Concerned with the concept of an integral, which measures accumulation, like area under a curve.

This exercise involves limits and differentiation, both vital components of calculus. By evaluating indeterminate forms using l'Hôpital's Rule, students practice applying differentiation to solve complex limit problems.

Indeterminate Forms

Indeterminate forms occur frequently in calculus when solving limits. These forms are not clearly defined at first and need more analysis to resolve. Common indeterminate forms include \", such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and others.
In this exercise, as \(x\) approaches \(0^{+}\) both the numerator \(2 \ln(x)\) and the denominator \(\ln(2x)\) tend toward \(-\infty\). This results in the indeterminate form \(\frac{-\infty}{-\infty}\).

• This form suggests that the initial computation doesn't provide a clear outcome.
• To solve it, you can apply techniques like l'Hôpital's Rule.

Transforming indeterminate forms into more manageable expressions allows us to find limits effectively.

Limits

The limit of a function describes the behavior of that function as its input approaches a particular point. Limits are essential in defining derivatives and integrals.

• Limits analyze the trend of a function's output near the point of interest.
• They can be one-sided, such as \(\lim_{x \to 0^{+}}\), or two-sided, considering both directions simultaneously.

In this scenario, we focus on a one-sided limit, where \(x\) approaches \(0\) from the positive side. Understanding this concept helps students methodically tackle problems in calculus and achieve precise results.

Differentiation

Differentiation, a crucial part of calculus, is the process of finding a derivative. It measures how a function's output value changes as its input value changes.

• Derivatives provide a way to calculate curves' slope at any point.
• They are essential in optimization, helping find maximum and minimum values of functions.

To apply l'Hôpital's Rule, you must differentiate both the numerator \(2 \ln(x)\) and the denominator \(\ln(2x)\). By simplifying these derivatives, you can evaluate the complex limit problem effectively.