For real numbers $ x$, let $ f(x)$ be the least integer that is greater than or equal to $ x$. Let $ a,\ b$ be positive real numbers, find the maximum value of $ c$ for which $ \lim_{x\to\infty} x^c\left(\frac{1}{f(ax\minus{}7)}\minus{}\frac{1}{f(bx\plus{}3)}\right)$ coverges, then find the limit value.
1. **Understanding the problem**: We need to find the maximum value of \( c \) for which the limit 
   \[
   \lim_{x \to \infty} x^c \left( \frac{1}{f(ax - 7)} - \frac{1}{f(bx + 3)} \right)
   \]
   converges, and then find the limit value.

2. **Using the definition of the ceiling function**: Recall that \( f(x) = \lceil x \rceil \), which is the smallest integer greater than or equal to \( x \). For large \( x \), we have:
   \[
   ax - 7 \le \lceil ax - 7 \rceil < ax - 6
   \]
   and
   \[
   bx + 3 \le \lceil bx + 3 \rceil < bx + 4
   \]

3. **Bounding the expression**: We can bound the given expression as follows:
   \[
   x^c \left( \frac{1}{ax - 6} - \frac{1}{bx + 4} \right) \le x^c \left( \frac{1}{\lceil ax - 7 \rceil} - \frac{1}{\lceil bx + 3 \rceil} \right) \le x^c \left( \frac{1}{ax - 7} - \frac{1}{bx + 3} \right)
   \]

4. **Simplifying the bounds**: For large \( x \), the differences between \( ax - 7 \) and \( ax - 6 \), and between \( bx + 3 \) and \( bx + 4 \), become negligible. Thus, we can approximate:
   \[
   x^c \left( \frac{1}{ax - 7} - \frac{1}{bx + 4} \right) \approx x^c \left( \frac{1}{ax} - \frac{1}{bx} \right)
   \]
   \[
   = x^c \left( \frac{b - a}{abx} \right)
   \]
   \[
   = \frac{b - a}{ab} x^{c-1}
   \]

5. **Finding the maximum \( c \)**: For the limit to converge, the exponent of \( x \) must be non-positive. Therefore, we require:
   \[
   c - 1 \le 0 \implies c \le 1
   \]
   The maximum value of \( c \) is thus \( 1 \).

6. **Calculating the limit**: When \( c = 1 \), the expression simplifies to:
   \[
   \lim_{x \to \infty} x \left( \frac{1}{\lceil ax - 7 \rceil} - \frac{1}{\lceil bx + 3 \rceil} \right)
   \]
   Using the bounds again, we get:
   \[
   \lim_{x \to \infty} x \left( \frac{1}{ax - 7} - \frac{1}{bx + 4} \right) = \lim_{x \to \infty} \frac{b - a}{ab} = \frac{b - a}{ab}
   \]

7. **Special case when \( a = b \)**: If \( a = b \), the expression inside the limit becomes:
   \[
   x^c \left( \frac{1}{\lceil ax - 7 \rceil} - \frac{1}{\lceil ax + 3 \rceil} \right)
   \]
   For large \( x \), this can be approximated as:
   \[
   x^c \left( \frac{1}{ax - 7} - \frac{1}{ax + 3} \right) = x^c \left( \frac{10}{(ax - 7)(ax + 3)} \right) \approx x^c \left( \frac{10}{a^2 x^2} \right) = \frac{10}{a^2} x^{c-2}
   \]
   For the limit to converge, we need \( c - 2 \le 0 \implies c \le 2 \). The maximum value of \( c \) is \( 2 \), and the limit is:
   \[
   \frac{10}{a^2}
   \]

The final answer is \( \boxed{ \frac{b - a}{ab} } \) if \( b \neq a \), and \( \frac{10}{a^2} \) if \( b = a \).