`lim_(n to oo) sum_(r=1)^(n) (1)/(n)e^(r//n)` is

Options

A. e+1
B. e-1
C. 1-e
D. e

Explanation

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}}, \]

we can interpret the sum as a Riemann sum, which approximates the integral of a function over an interval.

### Step-by-Step Solution:

1. **Identify the Riemann Sum**:
The expression \(\frac{1}{n} e^{\frac{r}{n}}\) can be seen as a Riemann sum for the function \(f(x) = e^x\) over the interval \([0, 1]\). Here, \(\frac{r}{n}\) represents a partition of the interval, where \(x = \frac{r}{n}\).

2. **Change of Variables**:
Let \(x = \frac{r}{n}\). As \(r\) goes from \(1\) to \(n\), \(x\) will go from \(\frac{1}{n}\) to \(1\). When \(n \to \infty\), \(\frac{1}{n} \to 0\). Thus, we can rewrite the limit as:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \lim_{n \to \infty} \sum_{r=1}^{n} \Delta x \cdot f(x), \]
where \(\Delta x = \frac{1}{n}\).

3. **Set Up the Integral**:
The limit of the Riemann sum as \(n\) approaches infinity gives us the integral:
\[ \int_{0}^{1} e^x \, dx. \]

4. **Evaluate the Integral**:
We can compute the integral:
\[ \int e^x \, dx = e^x + C. \]
Therefore,
\[ \int_{0}^{1} e^x \, dx = e^1 - e^0 = e - 1. \]

5. **Final Result**:
Thus, we have:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} = e - 1. \]

### Conclusion:
The final answer is

\[ \boxed{e - 1}. \]