Problem 16 - Thomas Calculus

Question

(Continution of Exercise \(15 . )\) Show that the value of $$ \lim _{x \rightarrow \infty} \frac{\ln (x+a)}{\ln x} $$ is the same no matter what value you assign to the constant \(a .\) What does this say about the relative rates at which the functions \(f(x)=\ln (x+a)\) and \(g(x)=\ln x\) grow?

Step-by-Step Solution

Answer

The limit is 1, indicating that \( \ln(x+a) \) and \( \ln x \) grow at the same rate as \( x \rightarrow \infty \).

Step 1: Express the Limit

Start with the given limit expression: \( \lim_{x \rightarrow \infty} \frac{\ln(x+a)}{\ln x} \). Our task is to show that this limit is the same for any constant \(a\).

Step 2: Apply Logarithmic Properties

Use the property of logarithms: \( \ln(x+a) = \ln(x(1+\frac{a}{x})) = \ln x + \ln(1+\frac{a}{x}) \). Substitute this expression into the limit, resulting in \( \frac{\ln x + \ln(1+\frac{a}{x})}{\ln x} \).

Step 3: Simplify the Expression

Divide each term in the numerator by \( \ln x \): \( \frac{\ln x}{\ln x} + \frac{\ln(1+\frac{a}{x})}{\ln x} = 1 + \frac{\ln(1+\frac{a}{x})}{\ln x} \). The limit now becomes \( \lim_{x \rightarrow \infty} \left( 1 + \frac{\ln(1+\frac{a}{x})}{\ln x} \right) \).

Step 4: Evaluate the Limit

As \( x \rightarrow \infty \), the term \( \frac{a}{x} \rightarrow 0 \). Hence, \( \ln(1+\frac{a}{x}) \approx \frac{a}{x} \) for large x. Thus, \( \frac{\ln(1+\frac{a}{x})}{\ln x} \approx \frac{\frac{a}{x}}{\ln x} \rightarrow 0 \) as \( x \rightarrow \infty \).

Step 5: Final Result

The limit expression simplifies to \( 1 + 0 = 1 \). Therefore, \( \lim_{x \rightarrow \infty} \frac{\ln(x+a)}{\ln x} = 1 \) independent of \(a\). This shows that \(f(x) = \ln(x+a)\) and \(g(x) = \ln x\) grow at the same rate at infinity.

Key Concepts

Limits at Infinity

The concept of 'limits at infinity' is crucial in understanding how functions behave as their input grows very large. Specifically, limits at infinity explore how a function behaves as the variable approaches a very large value, often denoted by \( x \rightarrow \infty \). For example, when exploring the limit \( \lim_{x \rightarrow \infty} \frac{\ln(x+a)}{\ln x} \), we are investigating how the ratio of these logarithmic functions changes as \( x \) becomes very large.
To solve such limits, we often simplify the expression using properties of logarithms and algebraic manipulation. In many cases, as \( x \) grows, certain terms may become negligible, simplifying the overall expression. Understanding limits at infinity allows us to determine the long-term behavior of functions, which is especially useful in calculus. Through the problem example, we found that the limit equaled 1, indicating that the two logarithmic functions grow at the same rate as \( x \rightarrow \infty \).

Properties of Logarithms

Logarithm properties are powerful tools that help simplify complex expressions. These properties include rules like \( \ln(ab) = \ln a + \ln b \) and \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \). In our exercise, we used the property \( \ln(x+a) = \ln(x(1+\frac{a}{x})) = \ln x + \ln(1+\frac{a}{x}) \) to break down and simplify the original logarithm expression.
This step is essential because it transforms a complicated limit expression into a more manageable form, which can then be analyzed as \( x \rightarrow \infty \). When working with limits involving logarithms, employing these properties simplifies the calculations and exposes hidden relationships between terms. This strategy of decisive simplification reveals the underlying growth behavior of the functions involved, as demonstrated in the solution, where the influence of \( a \) becomes negligible in the limit, leading us to the conclusion that the rate of growth is unaffected by the constant \( a \).

Growth Rates of Functions

Understanding growth rates of functions allows us to compare how rapidly different functions approach infinity. In mathematics, comparing growth rates becomes insightful when stemming from the function's structure or a limit, as observed when examining \( f(x) = \ln(x+a) \) and \( g(x) = \ln x \).
From the problem, we discovered that both functions share the same growth rate at infinity since the limit \( \lim_{x \rightarrow \infty} \frac{\ln(x+a)}{\ln x} = 1 \). It implies both \( f(x) \) and \( g(x) \) grow proportionally as \( x \) goes to infinity. Generally, growth rates let mathematicians classify functions into different orders of magnitude. This means that despite a shift resulting from \( a \), \( \ln(x+a) \) will not outgrow \( \ln x \) significantly or vice versa, as they both tend toward the same infinite growth.
Growth rates provide insights not only in theoretical exercises but also in practical scenarios, like computing resources and algorithms, where knowing which process will grow faster is critical.