Let $\{\epsilon_n\}^\infty_{n=1}$ be a sequence of positive reals with $\lim\limits_{n\rightarrow+\infty}\epsilon_n = 0$. Find \[ \lim\limits_{n\rightarrow\infty}\dfrac{1}{n}\sum\limits^{n}_{k=1}\ln\left(\dfrac{k}{n}+\epsilon_n\right)  \]
1. We start by considering the given limit:
   \[
   \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{k=1}^{n} \ln\left(\dfrac{k}{n} + \epsilon_n\right)
   \]
   Since \(\epsilon_n \to 0\) as \(n \to \infty\), we can think of the sum as a Riemann sum for an integral.

2. Rewrite the sum in a form that resembles a Riemann sum:
   \[
   \dfrac{1}{n} \sum_{k=1}^{n} \ln\left(\dfrac{k}{n} + \epsilon_n\right)
   \]
   This can be interpreted as the Riemann sum for the function \(f(x) = \ln(x + \epsilon_n)\) over the interval \([0, 1]\).

3. As \(n \to \infty\), the Riemann sum converges to the integral:
   \[
   \int_0^1 \ln(x + \epsilon_n) \, dx
   \]

4. We need to evaluate the integral:
   \[
   \int_0^1 \ln(x + \epsilon_n) \, dx
   \]

5. Use integration by parts to solve the integral. Let \(u = \ln(x + \epsilon_n)\) and \(dv = dx\). Then \(du = \frac{1}{x + \epsilon_n} dx\) and \(v = x\):
   \[
   \int \ln(x + \epsilon_n) \, dx = x \ln(x + \epsilon_n) - \int \frac{x}{x + \epsilon_n} \, dx
   \]

6. Simplify the second integral:
   \[
   \int \frac{x}{x + \epsilon_n} \, dx = \int \left(1 - \frac{\epsilon_n}{x + \epsilon_n}\right) \, dx = x - \epsilon_n \ln(x + \epsilon_n)
   \]

7. Combine the results:
   \[
   \int_0^1 \ln(x + \epsilon_n) \, dx = \left[ x \ln(x + \epsilon_n) - (x - \epsilon_n \ln(x + \epsilon_n)) \right]_0^1
   \]
   \[
   = \left[ \epsilon_n \ln(x + \epsilon_n) \right]_0^1
   \]
   \[
   = \epsilon_n \ln(1 + \epsilon_n) - \epsilon_n \ln(\epsilon_n)
   \]

8. Simplify the expression:
   \[
   = \epsilon_n \ln(1 + \epsilon_n) - \epsilon_n \ln(\epsilon_n)
   \]

9. As \(\epsilon_n \to 0\), use the approximation \(\ln(1 + \epsilon_n) \approx \epsilon_n\):
   \[
   \epsilon_n \ln(1 + \epsilon_n) \approx \epsilon_n \cdot \epsilon_n = \epsilon_n^2
   \]
   \[
   \epsilon_n \ln(\epsilon_n) \to 0
   \]

10. Therefore, the integral simplifies to:
    \[
    \int_0^1 \ln(x + \epsilon_n) \, dx \to -1
    \]

11. Hence, the original limit is:
    \[
    \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{k=1}^{n} \ln\left(\dfrac{k}{n} + \epsilon_n\right) = -1
    \]

The final answer is \( \boxed{ -1 } \)