\( \lim _ { n \rightarrow \infty } \sum _ { i = 1 } ^ { m } ( \frac { 1 } { m } ) [ 8 ( \frac { i } { n } + 3 ] \)

To evaluate the limit: \[ \lim_{n \rightarrow \infty} \sum_{i=1}^{m} \left( \frac{1}{m} \right) \left[ 8 \left( \frac{i}{n} + 3 \right) \right], \] let's break it down step by step. 1. **Simplify the Expression Inside the Sum:** \[ \frac{1}{m} \times 8 \left( \frac{i}{n} + 3 \right) = \frac{8}{m} \left( \frac{i}{n} + 3 \right) \] 2. **Distribute the Summation:** \[ \lim_{n \rightarrow \infty} \frac{8}{m} \sum_{i=1}^{m} \left( \frac{i}{n} + 3 \right) = \frac{8}{m} \left( \sum_{i=1}^{m} \frac{i}{n} + \sum_{i=1}^{m} 3 \right) \] 3. **Evaluate Each Sum Separately:** - The first sum: \[ \sum_{i=1}^{m} \frac{i}{n} = \frac{1}{n} \sum_{i=1}^{m} i = \frac{1}{n} \cdot \frac{m(m + 1)}{2} \] As \( n \rightarrow \infty \), this term approaches 0 because \( \frac{m(m + 1)}{2n} \rightarrow 0 \). - The second sum: \[ \sum_{i=1}^{m} 3 = 3m \] 4. **Combine the Results:** \[ \frac{8}{m} \left( 0 + 3m \right) = \frac{8}{m} \times 3m = 24 \] Therefore, the limit is: \[ \boxed{24} \]