To solve the given limit,

$\lim_{n \to \infty} \frac{\ln(2 + n)}{n + 1},$

let us proceed step by step.

Step 1: Analyze the behavior of the numerator and denominator

• The numerator is $\ln(2 + n)$, which grows without bound as $n \to \infty$, but much slower compared to linear or polynomial functions.
• The denominator is $n + 1$, which also grows without bound as $n \to \infty$, but linearly.

Step 2: Apply asymptotic approximation

For large $n$, $2 + n \approx n$. Thus, $\ln(2 + n) \approx \ln n$. The expression simplifies to:

$\lim_{n \to \infty} \frac{\ln(2 + n)}{n + 1} \approx \lim_{n \to \infty} \frac{\ln n}{n + 1}.$

Step 3: Simplify further

Since $n + 1 \approx n$ for large $n$, the expression becomes:

$\lim_{n \to \infty} \frac{\ln n}{n}.$

Step 4: Apply L’Hôpital’s Rule

The limit $\lim_{n \to \infty} \frac{\ln n}{n}$ is an indeterminate form ($\frac{\infty}{\infty}$). Differentiating the numerator and denominator gives:

$\lim_{n \to \infty} \frac{\ln n}{n} = \lim_{n \to \infty} \frac{\frac{d}{dn}(\ln n)}{\frac{d}{dn}(n)} = \lim_{n \to \infty} \frac{\frac{1}{n}}{1} = \lim_{n \to \infty} \frac{1}{n}.$

Step 5: Evaluate the result

$\lim_{n \to \infty} \frac{1}{n} = 0.$

Thus, the given limit evaluates to:

$\boxed{0}.$

Do you want a detailed explanation of any step? Here are 5 related questions you can explore:

1. How does L’Hôpital’s Rule work in solving indeterminate forms?
2. What are the growth rates of logarithmic versus polynomial functions?
3. Can you derive the approximation $\ln(2 + n) \approx \ln n$ for large $n$?
4. What are some other examples of limits involving logarithmic growth?
5. Why does $\frac{\ln n}{n}$ approach $0$ as $n \to \infty$?

Tip: When dealing with limits involving logarithms and polynomials, compare their growth rates to simplify the problem efficiently.