\( \lim _ { n \rightarrow \infty } \frac { 2 } { n ^ { 4 } } ( 1 + 8 + 27 + \ldots + n ^ { 3 } ) \)

To solve the limit \( \lim_{n \rightarrow \infty} \frac{2}{n^4} (1 + 8 + 27 + \ldots + n^3) \), we first need to calculate the sum inside the limit. The series \( 1 + 8 + 27 + \ldots + n^3 \) can be expressed as the sum of cubes: \[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \] Now we substitute this into our limit: \[ \lim_{n \rightarrow \infty} \frac{2}{n^4} \left( \frac{n(n+1)}{2} \right)^2 \] Calculating the square: \[ \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{4} = \frac{n^2(n^2 + 2n + 1)}{4} \] Thus, we have: \[ \lim_{n \rightarrow \infty} \frac{2}{n^4} \cdot \frac{n^2(n^2 + 2n + 1)}{4} = \lim_{n \rightarrow \infty} \frac{2n^2(n^2 + 2n + 1)}{4n^4} = \lim_{n \rightarrow \infty} \frac{n^2(n^2 + 2n + 1)}{2n^4} \] This simplifies to: \[ \lim_{n \rightarrow \infty} \frac{n^2 + 2n + 1}{2n^2} \] Breaking it down further, we have: \[ \frac{n^2}{2n^2} + \frac{2n}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{n} + \frac{1}{2n^2} \] Taking the limit as \( n \) approaches infinity yields: \[ \frac{1}{2} + 0 + 0 = \frac{1}{2} \] Therefore, the limit is: \[ \boxed{\frac{1}{2}} \]