Problem 31 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x}$$

Step-by-Step Solution

Answer

The limit is \( \ln(2) \).

Step 1: Identify the Indeterminate Form

Examine the given limit: \( \lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x} \). As \( x \rightarrow \infty \), both the numerator \( \ln(x+1) \) and the denominator \( \log_{2} x \) go to infinity. This is an indeterminate form of type \( \frac{\infty}{\infty} \).

Step 2: Differentiate the Numerator and Denominator

Apply L'Hôpital's Rule which allows us to differentiate the numerator and the denominator. The derivative of the numerator \( \ln(x+1) \) is \( \frac{1}{x+1} \). The derivative of the denominator \( \log_{2} x \) is \( \frac{1}{x \cdot \ln(2)} \) using the change of base formula.

Step 3: Apply L'Hôpital's Rule to the Derivatives

Use L'Hôpital's Rule: different derivatives lead to \( \lim_{x \to \infty} \frac{\ln (x+1)}{\log _{2} x} = \lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x \cdot \ln(2)}} \). Simplify this to \( \lim_{x \to \infty} \frac{x \cdot \ln(2)}{x+1} \).

Step 4: Simplify and Evaluate the Limit

Simplify the expression: \( \frac{x \cdot \ln(2)}{x+1} = \ln(2) \cdot \frac{x}{x+1} \). As \( x \rightarrow \infty \), \( \frac{x}{x+1} \rightarrow 1 \). Therefore, the limit becomes \( \ln(2) \cdot 1 = \ln(2) \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms often arise when evaluating limits that initially do not provide clear outcomes. These forms usually appear when limits involve expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In our problem, the limit \( \lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x} \) results in \( \frac{\infty}{\infty} \), which is an indeterminate form. This means that as \( x \) approaches infinity, both the numerator and the denominator grow indefinitely.

Examples of indeterminate forms include:

• \( \frac{0}{0} \) - Often occurs when both the function's numerator and denominator approach zero.
• \( \frac{\infty}{\infty} \) - Shows up when both the numerator and denominator grow infinitely large.
• Other forms like \( 0 \times \infty \), \( \infty - \infty \), and more.

Understanding these conditions is the key step in applying techniques like L'Hôpital's Rule to resolve these uncertain outcomes in calculus.

Limits in Calculus

Limits are a fundamental concept in calculus that help us understand the behavior of functions as values approach a specific point. When we encounter a problem like \( \lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x} \), we're tasked to discover what value, if any, the function approaches as \( x \) goes to infinity.

The technique of taking limits is crucial for:

• Defining derivatives, which describe how functions change.
• Establishing integrals, which calculate areas under curves.
• Solving problems involving convergence of series and other mathematical constructs.

Understanding limits is essential as it lays the groundwork for more advanced topics in calculus and analysis. L'Hôpital's Rule is a specific tool designed to handle limits with indeterminate forms by using derivatives, making complex limit problems solvable.

Logarithmic Functions

Logarithmic functions, like those involving the natural log \( \ln(x) \) and the logarithm base 2 \( \log_{2}x \), play a significant role in calculus. In our exercise, we need to find the limit involving these functions: \( \ln(x+1) \) and \( \log_{2}x \).

Some important characteristics of logarithmic functions include:

• The natural logarithm, \( \ln(x) \), is the inverse of the exponential function \( e^x \).
• Logarithmic functions grow slowly compared to polynomial or exponential functions, which is why they often relate to growth processes.
• Using the change of base formula, \( \log_{b} a = \frac{\ln a}{\ln b} \), logarithms of different bases can be converted for easier manipulation.

In the problem at hand, understanding the derivatives of these functions is key. Calculating derivatives of \( \ln(x+1) \) and \( \log_2(x) \) allowed for the application of L'Hôpital's Rule to resolve the indeterminate form and find the limit. Thus, logarithmic functions are pivotal in aiding the transformation and simplification of complex calculus problems.