To solve the problem, we need to find the limit of the ratio of the nth terms of two sequences as n approaches infinity. Let's denote the nth term of the first sequence as \( f(n) \) and the nth term of the second sequence as \( g(n) \).
### Step 1: Identify the nth terms of the sequences
The first sequence is \( 3, 6, 11, 18, 27, \ldots \). We can observe the differences between consecutive terms:
- \( 6 - 3 = 3 \)
- \( 11 - 6 = 5 \)
- \( 18 - 11 = 7 \)
- \( 27 - 18 = 9 \)
The differences are \( 3, 5, 7, 9, \ldots \), which form an arithmetic sequence with a common difference of 2.
Thus, we can express the nth term \( f(n) \) as:
\[
f(n) = n^2 + 2
\]
The second sequence is \( 3, 7, 13, 21, \ldots \). The differences between consecutive terms are:
- \( 7 - 3 = 4 \)
- \( 13 - 7 = 6 \)
- \( 21 - 13 = 8 \)
The differences are \( 4, 6, 8, \ldots \), which also form an arithmetic sequence with a common difference of 2.
Thus, we can express the nth term \( g(n) \) as:
\[
g(n) = n^2 + n + 1
\]
### Step 2: Set up the limit
We need to find:
\[
\lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{n^2 + 2}{n^2 + n + 1}
\]
### Step 3: Simplify the limit
To simplify this limit, we divide both the numerator and the denominator by \( n^2 \):
\[
\lim_{n \to \infty} \frac{1 + \frac{2}{n^2}}{1 + \frac{1}{n} + \frac{1}{n^2}}
\]
### Step 4: Evaluate the limit
As \( n \) approaches infinity, the terms \( \frac{2}{n^2} \), \( \frac{1}{n} \), and \( \frac{1}{n^2} \) all approach 0. Therefore, the limit simplifies to:
\[
\frac{1 + 0}{1 + 0 + 0} = \frac{1}{1} = 1
\]
### Final Answer
Thus, the limit is:
\[
\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1
\]
