For positive real numbers $ a,\ b$, two graphs of the function : $ x^a$ and $ \ln bx$ have a tangency of point.

(1) Let $ (s,\ t)$ be the tangency of point, express this in terms of $ a$, then express $ b$ as the function of $ a$.

(2) For $ h$ with $ 0 < h < s$,  denote the area as $ A(h)$ of the domain bounded by the line $ x \equal{} h$ and two curves $ y \equal{} x^a,\ y \equal{} \ln bx$.

Express $ \lim_{h\rightarrow 0} A(h)$ in terms of $ a$.
### Part 1: Finding the Tangency Point and Expressing \( b \) as a Function of \( a \)

1. **Equating the derivatives:**
   To find the tangency point, we need to equate the derivatives of the functions \( y = x^a \) and \( y = \ln(bx) \).

   \[
   \frac{d}{dx}(x^a) = \frac{d}{dx}(\ln(bx))
   \]

   The derivative of \( x^a \) is:

   \[
   \frac{d}{dx}(x^a) = a x^{a-1}
   \]

   The derivative of \( \ln(bx) \) is:

   \[
   \frac{d}{dx}(\ln(bx)) = \frac{1}{bx} \cdot b = \frac{1}{x}
   \]

   Setting these equal gives:

   \[
   a x^{a-1} = \frac{1}{x}
   \]

2. **Solving for \( x \):**

   \[
   a x^{a-1} = \frac{1}{x} \implies a x^a = 1 \implies x = a^{-\frac{1}{a}}
   \]

   So, the \( x \)-coordinate of the tangency point is \( s = a^{-\frac{1}{a}} \).

3. **Finding the \( y \)-coordinate:**

   Substitute \( x = a^{-\frac{1}{a}} \) into either function to find the \( y \)-coordinate. Using \( y = x^a \):

   \[
   t = \left(a^{-\frac{1}{a}}\right)^a = a^{-1}
   \]

   Thus, the tangency point is \( (s, t) = \left(a^{-\frac{1}{a}}, a^{-1}\right) \).

4. **Expressing \( b \) as a function of \( a \):**

   At the tangency point, the functions must have the same value:

   \[
   \left(a^{-\frac{1}{a}}\right)^a = \ln\left(b a^{-\frac{1}{a}}\right)
   \]

   Simplifying the left side:

   \[
   a^{-1} = \ln\left(b a^{-\frac{1}{a}}\right)
   \]

   Solving for \( b \):

   \[
   a^{-1} = \ln(b) + \ln\left(a^{-\frac{1}{a}}\right) = \ln(b) - \frac{1}{a} \ln(a)
   \]

   \[
   a^{-1} + \frac{1}{a} \ln(a) = \ln(b)
   \]

   \[
   \ln(b) = a^{-1} + \frac{1}{a} \ln(a) = \frac{1 + \ln(a)}{a}
   \]

   \[
   b = e^{\frac{1 + \ln(a)}{a}} = (ae)^{\frac{1}{a}}
   \]

### Part 2: Finding the Limit of the Area \( A(h) \)

1. **Setting up the integral:**

   The area \( A(h) \) is given by the integral:

   \[
   A(h) = \int_h^{a^{-\frac{1}{a}}} \left( x^a - \ln\left((ae)^{\frac{1}{a}} x \right) \right) \, dx
   \]

2. **Simplifying the integrand:**

   \[
   \ln\left((ae)^{\frac{1}{a}} x \right) = \ln\left((ae)^{\frac{1}{a}}\right) + \ln(x) = \frac{1 + \ln(a)}{a} + \ln(x)
   \]

   So the integrand becomes:

   \[
   x^a - \left( \frac{1 + \ln(a)}{a} + \ln(x) \right) = x^a - \frac{1 + \ln(a)}{a} - \ln(x)
   \]

3. **Evaluating the integral:**

   \[
   A(h) = \int_h^{a^{-\frac{1}{a}}} \left( x^a - \frac{1 + \ln(a)}{a} - \ln(x) \right) \, dx
   \]

   This integral can be split into three parts:

   \[
   A(h) = \int_h^{a^{-\frac{1}{a}}} x^a \, dx - \int_h^{a^{-\frac{1}{a}}} \frac{1 + \ln(a)}{a} \, dx - \int_h^{a^{-\frac{1}{a}}} \ln(x) \, dx
   \]

   Evaluating each part separately:

   \[
   \int_h^{a^{-\frac{1}{a}}} x^a \, dx = \left[ \frac{x^{a+1}}{a+1} \right]_h^{a^{-\frac{1}{a}}} = \frac{(a^{-\frac{1}{a}})^{a+1}}{a+1} - \frac{h^{a+1}}{a+1}
   \]

   \[
   \int_h^{a^{-\frac{1}{a}}} \frac{1 + \ln(a)}{a} \, dx = \frac{1 + \ln(a)}{a} \left[ x \right]_h^{a^{-\frac{1}{a}}} = \frac{1 + \ln(a)}{a} \left( a^{-\frac{1}{a}} - h \right)
   \]

   \[
   \int_h^{a^{-\frac{1}{a}}} \ln(x) \, dx = \left[ x \ln(x) - x \right]_h^{a^{-\frac{1}{a}}} = \left( a^{-\frac{1}{a}} \ln(a^{-\frac{1}{a}}) - a^{-\frac{1}{a}} \right) - \left( h \ln(h) - h \right)
   \]

   Simplifying:

   \[
   a^{-\frac{1}{a}} \ln(a^{-\frac{1}{a}}) = a^{-\frac{1}{a}} \left( -\frac{\ln(a)}{a} \right) = -\frac{\ln(a)}{a} a^{-\frac{1}{a}}
   \]

   So:

   \[
   \int_h^{a^{-\frac{1}{a}}} \ln(x) \, dx = -\frac{\ln(a)}{a} a^{-\frac{1}{a}} - a^{-\frac{1}{a}} - (h \ln(h) - h)
   \]

4. **Combining the results:**

   \[
   A(h) = \frac{a^{-\frac{a-1}{a}}}{a+1} - \frac{h^{a+1}}{a+1} - \frac{1 + \ln(a)}{a} \left( a^{-\frac{1}{a}} - h \right) - \left( -\frac{\ln(a)}{a} a^{-\frac{1}{a}} - a^{-\frac{1}{a}} - (h \ln(h) - h) \right)
   \]

5. **Taking the limit as \( h \to 0 \):**

   As \( h \to 0 \):

   \[
   \lim_{h \to 0} A(h) = \frac{a^{-\frac{a-1}{a}}}{a+1} - 0 - \frac{1 + \ln(a)}{a} a^{-\frac{1}{a}} - \left( -\frac{\ln(a)}{a} a^{-\frac{1}{a}} - a^{-\frac{1}{a}} \right)
   \]

   Simplifying:

   \[
   \lim_{h \to 0} A(h) = \frac{a^{-\frac{a-1}{a}}}{a+1}
   \]

The final answer is \( \boxed{ \frac{a^{-\frac{a-1}{a}}}{a+1} } \)