Problem 76 - Essential Calculus Early Transcendentals

Question

Find the limit. $$ \lim _{x \rightarrow \infty}[\ln (2+x)-\ln (1+x)] $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the problem

We are given a limit problem where the expression is the difference of two natural logarithms. To find \( \lim _{x \rightarrow \infty} [\ln(2+x) - \ln(1+x)] \), we will use the properties of logarithms.

Step 2: Apply Logarithmic Properties

Use the logarithmic property that states \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). Applying this to our expression, we get \( \ln\left(\frac{2+x}{1+x}\right) \).

Step 3: Simplify the Expression Inside the Logarithm

Inside the logarithm, simplify \( \frac{2+x}{1+x} \) by dividing both the numerator and the denominator by \( x \). This results in \( \frac{\frac{2}{x} + 1}{\frac{1}{x} + 1} \).

Step 4: Evaluate the Limit as \( x \to \infty \)

As \( x \to \infty \), both \( \frac{2}{x} \) and \( \frac{1}{x} \) approach 0, so the expression simplifies to \( \frac{0+1}{0+1} = 1 \). Thus, we evaluate \( \ln(1) \), which is 0.

Step 5: Confirm the Result

By considering the behavior of the original expression as \( x \to \infty \), we confirm that \( \lim _{x \rightarrow \infty} \ln\left(\frac{2+x}{1+x}\right) = \ln(1) = 0 \).

Key Concepts

Logarithmic Properties

Logarithms might seem daunting, but they have handy properties that can simplify complex expressions. One key property is the difference of logarithms: given two positive numbers, \(a\) and \(b\), the property states \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This helps reduce a subtraction of logs into a single log expression.

This property is crucial in calculus because it makes the evaluation of limits and derivatives involving logarithms more manageable. By transforming two separate logs into a single one, it's easier to identify patterns, like approaching limits or recognizing indeterminate forms.

For instance, in our exercise, we used this property to simplify \( \ln(2+x) - \ln(1+x) \) into \( \ln\left(\frac{2+x}{1+x}\right) \). This step is pivotal because it sets the stage for further simplification and eventual limit evaluation.

Simplification Techniques

Simplifying expressions in calculus helps reveal the behavior of functions as variables change. In the given limit exercise, another useful technique involved rational expressions and simplification through division.

With \( \frac{2+x}{1+x} \), we can simplify by dividing the numerator and the denominator both by \(x\). This common technique aims to expose how the expression behaves as \(x\) grows to infinity. After dividing, we obtain \( \frac{\frac{2}{x} + 1}{\frac{1}{x} + 1} \).

This strategy is efficient because as \(x\) heads towards infinity, terms like \(\frac{2}{x}\) and \(\frac{1}{x}\) tend to zero. By simplifying the form of the fraction, it becomes easier to see which terms "matter" as \(x\) becomes very large. Such simplifications are mainstream in calculus and aid significantly in limit evaluation.

Infinity in Calculus

Infinity often appears in calculus, particularly when examining limits. When \(x\) approaches infinity, we are interested in observing the behavior of functions as they stretch towards endlessly large values.

In the exercise, we calculated the limit as \(x\) approaches infinity. Our goal is to see how the expression \( \ln\left(\frac{2+x}{1+x}\right) \) behaves when \(x\) is very large. Using our simplification, terms such as \(\frac{2}{x}\) become negligible because they tend to zero, thus allowing the fraction to effectively reduce to \(\frac{1}{1} = 1\).

Understanding infinity in calculus doesn't mean grasping an abstract idea of endlessness. Rather, it's about recognizing how expressions behave or stabilize as they head towards large values. By simplifying and reducing the expression as we did, you can see that \( \ln(1) = 0 \), thus concluding this limit problem efficiently.