Problem 15 - Thomas Calculus

Question

Investigate $$ \lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\ln x} \quad \text { and } \quad \lim _{x \rightarrow \infty} \frac{\ln (x+999)}{\ln x} $$ Then use l'Hôpital's Rule to explain what you find.

Step-by-Step Solution

Answer

Both limits are 1, indicating similar growth rates for \( \ln(x+1) \) and \( \ln(x+999) \) compared to \( \ln x \).

Step 1: Identify the Type of Limit Problem

Both expressions \( \lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\ln x} \) and \( \lim _{x \rightarrow \infty} \frac{\ln (x+999)}{\ln x} \) are of the form \( \frac{\infty}{\infty} \), which allows us to use l'Hôpital's Rule. The approach involves differentiating the numerator and the denominator separately.

Step 2: Differentiate Numerator and Denominator

Differentiate the functions in both limits:For \( f(x) = \ln(x+1) \), its derivative is \( f'(x) = \frac{1}{x+1} \).For \( g(x) = \ln(x+999) \), its derivative is \( g'(x) = \frac{1}{x+999} \).For the denominator \( h(x) = \ln(x) \), its derivative is \( h'(x) = \frac{1}{x} \).

Step 3: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule to the first limit:\[ \lim_{x \rightarrow \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{x}{x+1} \]This simplifies to \( \lim_{x \rightarrow \infty} \left( 1 - \frac{1}{x+1} \right) = 1 \).Similarly, for the second limit:\[ \lim_{x \rightarrow \infty} \frac{\frac{1}{x+999}}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{x}{x+999} \]This simplifies to \( \lim_{x \rightarrow \infty} \left( 1 - \frac{999}{x+999} \right) = 1 \).

Step 4: Conclusion from Limits

Both limits \( \lim _{x \rightarrow \infty} \frac{\ln(x+1)}{\ln x} \) and \( \lim _{x \rightarrow \infty} \frac{\ln(x+999)}{\ln x} \), upon applying l'Hôpital's Rule, approach the value 1 as \( x \rightarrow \infty \). This indicates that as \( x \) becomes very large, the logarithm of \( x+1 \) and \( x+999 \) grow at the same rate as \( \ln x \).

Key Concepts

Understanding Limits

When dealing with mathematical problems involving limits, it's essential to understand what this concept means. Essentially, a limit describes the value that a function approaches as the input approaches a certain point. In the context of the original exercise, we are interested in what happens to the expressions as \( x \) approaches infinity.

• The expressions are of the form \( \frac{\ln(x+1)}{\ln x} \) and \( \frac{\ln(x+999)}{\ln x} \).
• Both are considered indeterminate forms of type \( \frac{\infty}{\infty} \), which is a key point for using certain mathematical techniques.

The goal here is to determine whether these ratios converge to a certain number or pattern when \( x \) gets extremely large. Limits help us to understand the behavior of functions at the boundaries and can tell us how similar different functions are in their growth rates.

Exploring Logarithmic Functions

Logarithmic functions are a crucial part of the problem, specifically natural logarithms, denoted as \( \ln(x) \). Logarithms answer the question: "To what power must the base (usually \( e \), the base of natural logarithms) be raised to yield the input number?" Here's what to know about them:

• They grow slower compared to linear or polynomial functions, which is why they are compelling for examining growth rates.
• When we look at \( \ln(x+1) \) and \( \ln(x+999) \), we see that as \( x \) becomes very large, adding 1 or even 999 becomes negligible in the scale of the growth of \( x \).

With logarithmic functions, small additions to \( x \) hardly affect the growth when \( x \) is very large. This explains why both limits in the original problem evaluate to 1. Essentially, \( \ln(x+1) \) and \( \ln(x+999) \) grow with the same order as \( \ln(x) \).

The Role of Derivatives in l'Hôpital's Rule

Derivatives are the mathematical tools that describe the rate at which a function changes. They are integral in using l'Hôpital's Rule, which is designed to solve indeterminate forms like \( \frac{\infty}{\infty} \). Here's a basic outline:

• We compute the derivatives: for \( \ln(x+1) \) it is \( \frac{1}{x+1} \), \( \ln(x+999) \) is \( \frac{1}{x+999} \), and \( \ln(x) \) is \( \frac{1}{x} \).
• l'Hôpital's Rule states that for indeterminate forms, the limit of a ratio of functions is equal to the limit of the ratio of their derivatives, if the limits exist.
• This converts our limits from a fraction of logarithms to a simple algebraic fraction: \( \frac{\frac{1}{x+1}}{\frac{1}{x}} \rightarrow \frac{x}{x+1} \) and similarly for the second expression.

Through differentiation, we transformed the problem into simpler fractions that were straightforward to evaluate as \( x \) approaches infinity. Hence, we determined both limits equal 1 when using l'Hôpital's Rule. This illustrates how derivatives help in understanding the behavior of complex functions and solving challenging limit problems.