`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 this expression as a Riemann sum, which can be converted into a definite integral as \( n \) approaches infinity.

### Step-by-step Solution:

1. **Identify the Riemann Sum**:
The expression \(\sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}}\) can be recognized as a Riemann sum for the function \(f(x) = e^x\) over the interval \([0, 1]\). Here, \(\frac{1}{n}\) represents the width of each subinterval, and \(e^{\frac{r}{n}}\) represents the function evaluated at the right endpoint of each subinterval.

2. **Set Up the Integral**:
As \(n\) approaches infinity, the sum converges to the integral of the function from 0 to 1:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \int_{0}^{1} e^x \, dx. \]

3. **Calculate the Integral**:
Now, we compute the integral:
\[ \int e^x \, dx = e^x + C. \]
Evaluating this from 0 to 1 gives:
\[ \int_{0}^{1} e^x \, dx = e^1 - e^0 = e - 1. \]

4. **Final Result**:
Therefore, the limit is:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} = e - 1. \]

### Conclusion:
The final answer is:
\[ e - 1. \]