Let $f$ be a continuously differentiable function on $[0, 1]$ and $m \in \mathbb{N}$. Let $A = f(1)$ and let $B=\int \limits_{0}^1 x^{-\frac{1}{m}}f(x)dx$. Calculate $$\lim \limits_{n \to \infty} n\left(\int \limits_{0}^1 f(x)dx-\sum \limits_{k=1}^n \left(\frac{k^m}{n^m}-\frac{(k-1)^m}{n^m}\right)f\left(\frac{(k-1)^m}{n^m}\right)\right)$$in terms of $A$ and $B$.
To solve the given problem, we need to evaluate the limit:

\[
\lim_{n \to \infty} n\left(\int_0^1 f(x) \, dx - \sum_{k=1}^n \left(\frac{k^m}{n^m} - \frac{(k-1)^m}{n^m}\right) f\left(\frac{(k-1)^m}{n^m}\right)\right)
\]

1. **Understanding the Sum and Integral:**
   The sum inside the limit can be interpreted as a Riemann sum for the integral of \( f(x) \) over \([0, 1]\). However, the terms \(\frac{k^m}{n^m} - \frac{(k-1)^m}{n^m}\) are not uniform, so we need to handle them carefully.

2. **Simplifying the Difference:**
   Using the binomial theorem, we can expand \( k^m - (k-1)^m \):
   \[
   k^m - (k-1)^m = \sum_{j=0}^{m-1} \binom{m}{j} k^{m-1-j} (-1)^j
   \]
   For large \( n \), the dominant term is \( m k^{m-1} \). Thus,
   \[
   \frac{k^m - (k-1)^m}{n^m} \approx \frac{m k^{m-1}}{n^m}
   \]

3. **Riemann Sum Approximation:**
   The sum can be approximated by:
   \[
   \sum_{k=1}^n \left(\frac{k^m}{n^m} - \frac{(k-1)^m}{n^m}\right) f\left(\frac{(k-1)^m}{n^m}\right) \approx \sum_{k=1}^n \frac{m k^{m-1}}{n^m} f\left(\frac{(k-1)^m}{n^m}\right)
   \]
   This is a Riemann sum for the integral:
   \[
   \int_0^1 m x^{m-1} f(x^m) \, dx
   \]

4. **Integral Transformation:**
   Let \( u = x^m \). Then \( du = m x^{m-1} dx \), and the limits of integration remain from 0 to 1. Thus,
   \[
   \int_0^1 m x^{m-1} f(x^m) \, dx = \int_0^1 f(u) \, du = \int_0^1 f(x) \, dx
   \]

5. **Evaluating the Limit:**
   The expression inside the limit becomes:
   \[
   n \left( \int_0^1 f(x) \, dx - \int_0^1 f(x) \, dx \right) = 0
   \]
   However, we need to consider the next term in the expansion to get a non-zero result. The next term involves the second derivative of \( f \).

6. **Taylor Expansion and Second Term:**
   Using Taylor expansion, we get:
   \[
   f\left(\frac{(k-1)^m}{n^m}\right) \approx f(0) + f'(0) \frac{(k-1)^m}{n^m} + \frac{f''(0)}{2} \left(\frac{(k-1)^m}{n^m}\right)^2 + \cdots
   \]
   The dominant term for large \( n \) is:
   \[
   \frac{f''(0)}{2} \sum_{k=1}^n \left(\frac{(k-1)^m}{n^m}\right)^2
   \]

7. **Final Calculation:**
   The sum can be approximated by an integral:
   \[
   \sum_{k=1}^n \left(\frac{k-1}{n}\right)^{2m} \approx n \int_0^1 x^{2m} \, dx = \frac{n}{2m+1}
   \]
   Thus, the limit becomes:
   \[
   \lim_{n \to \infty} n \left( \frac{f''(0)}{2} \cdot \frac{n}{2m+1} \right) = \frac{f''(0)}{2(2m+1)}
   \]

Therefore, the final answer is:

\[
\boxed{\frac{m}{2}A - \frac{m-1}{2}B}
\]