Let $C: y=\ln x$. For each positive integer $n$, denote by $A_n$ the area of the part enclosed by the line passing through two points $(n,\ \ln n),\ (n+1,\ \ln (n+1))$ and denote by $B_n$ that of the part enclosed by the tangent line at the point $(n,\ \ln n)$, $C$ and the line $x=n+1$. Let $g(x)=\ln (x+1)-\ln x$.

(1) Express $A_n,\ B_n$ in terms of $n,\ g(n)$ respectively.

(2) Find $\lim_{n\to\infty} n\{1-ng(n)\}$.
1. To express \(A_n\) and \(B_n\) in terms of \(n\) and \(g(n)\), we need to analyze the areas described in the problem.

   **Step 1: Express \(A_n\) in terms of \(n\) and \(g(n)\)**

   The area \(A_n\) is the area enclosed by the line passing through the points \((n, \ln n)\) and \((n+1, \ln (n+1))\), and the curve \(y = \ln x\).

   The line passing through these points can be found using the slope formula:
   \[
   m = \frac{\ln(n+1) - \ln n}{(n+1) - n} = \ln(n+1) - \ln n = g(n)
   \]
   The equation of the line is:
   \[
   y - \ln n = g(n)(x - n) \implies y = g(n)x + \ln n - ng(n)
   \]

   The area under the curve \(y = \ln x\) from \(n\) to \(n+1\) is:
   \[
   \int_n^{n+1} \ln x \, dx
   \]
   Using integration by parts, let \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} dx\) and \(v = x\):
   \[
   \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C
   \]
   Evaluating from \(n\) to \(n+1\):
   \[
   \int_n^{n+1} \ln x \, dx = \left[ x \ln x - x \right]_n^{n+1} = (n+1) \ln (n+1) - (n+1) - (n \ln n - n)
   \]
   \[
   = (n+1) \ln (n+1) - n \ln n - 1
   \]

   The area of the trapezoid formed by the line and the x-axis is:
   \[
   \frac{1}{2} \left( \ln n + \ln (n+1) \right) \cdot 1 = \frac{1}{2} (\ln n + \ln (n+1))
   \]

   Therefore, \(A_n\) is:
   \[
   A_n = \left( (n+1) \ln (n+1) - n \ln n - 1 \right) - \frac{1}{2} (\ln n + \ln (n+1))
   \]
   \[
   = (n+1) \ln (n+1) - n \ln n - 1 - \frac{1}{2} (\ln n + \ln (n+1))
   \]
   \[
   = (n + \frac{1}{2}) g(n) - 1
   \]

   **Step 2: Express \(B_n\) in terms of \(n\) and \(g(n)\)**

   The area \(B_n\) is the area enclosed by the tangent line at the point \((n, \ln n)\), the curve \(y = \ln x\), and the line \(x = n+1\).

   The tangent line at \((n, \ln n)\) has the slope:
   \[
   \frac{d}{dx} (\ln x) \bigg|_{x=n} = \frac{1}{n}
   \]
   The equation of the tangent line is:
   \[
   y - \ln n = \frac{1}{n} (x - n) \implies y = \frac{1}{n} x + \ln n - 1
   \]

   The area under the tangent line from \(n\) to \(n+1\) is:
   \[
   \int_n^{n+1} \left( \frac{1}{n} x + \ln n - 1 \right) dx
   \]
   \[
   = \left[ \frac{1}{2n} x^2 + (\ln n - 1)x \right]_n^{n+1}
   \]
   \[
   = \left( \frac{1}{2n} (n+1)^2 + (\ln n - 1)(n+1) \right) - \left( \frac{1}{2n} n^2 + (\ln n - 1)n \right)
   \]
   \[
   = \left( \frac{1}{2n} (n^2 + 2n + 1) + n \ln n + \ln n - n - 1 \right) - \left( \frac{1}{2} n + n \ln n - n \right)
   \]
   \[
   = \left( \frac{1}{2} n + 1 + \frac{1}{2n} + n \ln n + \ln n - n - 1 \right) - \left( \frac{1}{2} n + n \ln n - n \right)
   \]
   \[
   = \frac{1}{2n} + \ln n
   \]

   Therefore, \(B_n\) is:
   \[
   B_n = \left( \frac{1}{2n} + \ln n \right) - \left( (n+1) \ln (n+1) - n \ln n - 1 \right)
   \]
   \[
   = \frac{1}{2n} + \ln n - (n+1) g(n) + 1
   \]

2. To find \(\lim_{n \to \infty} n \{1 - n g(n)\}\), we need to analyze the behavior of \(g(n)\) as \(n \to \infty\).

   **Step 1: Analyze \(g(n)\) as \(n \to \infty\)**

   Recall that:
   \[
   g(n) = \ln(n+1) - \ln n = \ln \left( \frac{n+1}{n} \right) = \ln \left( 1 + \frac{1}{n} \right)
   \]

   Using the approximation \(\ln(1 + x) \approx x\) for small \(x\):
   \[
   g(n) \approx \frac{1}{n}
   \]

   **Step 2: Evaluate the limit**

   Substitute \(g(n) \approx \frac{1}{n}\) into the expression:
   \[
   \lim_{n \to \infty} n \left( 1 - n g(n) \right) = \lim_{n \to \infty} n \left( 1 - n \cdot \frac{1}{n} \right)
   \]
   \[
   = \lim_{n \to \infty} n (1 - 1) = \lim_{n \to \infty} n \cdot 0 = 0
   \]

The final answer is \(\boxed{0}\)