To solve the limit problem
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{j=0}^{n} \frac{(2j - 1) + 8n}{(2j - 1) + 4n},
\]
we will follow these steps:
### Step 1: Rewrite the expression
We start with the expression inside the limit:
\[
\frac{(2j - 1) + 8n}{(2j - 1) + 4n} = \frac{2j - 1 + 8n}{2j - 1 + 4n}.
\]
### Step 2: Simplify the fraction
To simplify the fraction, we can divide the numerator and the denominator by \(n\):
\[
= \frac{\frac{2j - 1}{n} + 8}{\frac{2j - 1}{n} + 4}.
\]
### Step 3: Substitute \(x = \frac{j}{n}\)
Now, we can express \(j\) in terms of \(n\) by letting \(x = \frac{j}{n}\). As \(j\) goes from \(0\) to \(n\), \(x\) will go from \(0\) to \(1\). The increment \(j\) can be expressed as \(j = nx\), and thus:
\[
\frac{2j - 1}{n} = 2x - \frac{1}{n}.
\]
As \(n \to \infty\), \(-\frac{1}{n} \to 0\). Therefore, we have:
\[
\frac{2j - 1 + 8n}{2j - 1 + 4n} \to \frac{2x + 8}{2x + 4}.
\]
### Step 4: Rewrite the sum as an integral
Now, we can rewrite the sum as an integral. The limit of the sum can be expressed as:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{j=0}^{n} \frac{(2j - 1) + 8n}{(2j - 1) + 4n} \to \int_{0}^{1} \frac{2x + 8}{2x + 4} \, dx.
\]
### Step 5: Evaluate the integral
Now we will evaluate the integral:
\[
\int_{0}^{1} \frac{2x + 8}{2x + 4} \, dx.
\]
To simplify, we can break it down:
\[
= \int_{0}^{1} \left(1 + \frac{4}{2x + 4}\right) dx.
\]
This gives us two integrals:
\[
= \int_{0}^{1} 1 \, dx + 4 \int_{0}^{1} \frac{1}{2x + 4} \, dx.
\]
The first integral evaluates to:
\[
\int_{0}^{1} 1 \, dx = 1.
\]
For the second integral, we can use the substitution \(u = 2x + 4\), which gives \(du = 2dx\) or \(dx = \frac{du}{2}\). The limits change from \(x = 0\) to \(x = 1\) which corresponds to \(u = 4\) to \(u = 6\):
\[
4 \int_{4}^{6} \frac{1}{u} \cdot \frac{du}{2} = 2 \int_{4}^{6} \frac{1}{u} \, du = 2 [\ln u]_{4}^{6} = 2 (\ln 6 - \ln 4) = 2 \ln \frac{6}{4} = 2 \ln \frac{3}{2}.
\]
### Step 6: Combine results
Combining both parts of the integral, we have:
\[
1 + 2 \ln \frac{3}{2}.
\]
### Final Result
Thus, the final result is:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{j=0}^{n} \frac{(2j - 1) + 8n}{(2j - 1) + 4n} = 1 + 2 \ln \frac{3}{2}.
\]
