Step-by-step Solution:

The function can be written as: \[ \lim_{n \to \infty} \frac{n^a \sum_{r=1}^n \left( \frac{r}{n} \right)^a}{(n+1)^{a-1} n \sum_{r=1}^n \left( a + \frac{r}{n} \right)} = \frac{1}{60}. \] Simplifying: \[ \Rightarrow \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^{a-1} \frac{\sum_{r=1}^n \left( \frac{r}{n} \right)^a}{\sum_{r=1}^n \left( a + \frac{r}{n} \right)}. \] As \( n \to \infty \), \( \frac{n}{n+1} \to 1 \), so: \[ \lim_{n \to \infty} \frac{\sum_{r=1}^n \left( \frac{r}{n} \right)^a}{\sum_{r=1}^n \left( a + \frac{r}{n} \right)} = \frac{\int_0^1 x^a \, dx}{\int_0^1 (a + x) \, dx}. \] Evaluating the integrals: \[ \int_0^1 x^a \, dx = \frac{1}{a+1}, \quad \int_0^1 (a + x) \, dx = \int_0^1 a \, dx + \int_0^1 x \, dx = a + \frac{1}{2}. \] Thus: \[ \frac{\int_0^1 x^a \, dx}{\int_0^1 (a + x) \, dx} = \frac{\frac{1}{a+1}}{a + \frac{1}{2}} = \frac{1}{(a+1)(2a+1)}. \] Given that: \[ \frac{2}{(a+1)(2a+1)} = \frac{1}{60}, \] we solve: \[ (2a+1)(a+1) = 120. \] Expanding: \[ 2a^2 + 3a + 1 = 120, \quad 2a^2 + 3a - 119 = 0. \] Solving this quadratic equation: \[ a = \frac{-3 \pm \sqrt{3^2 - 4(2)(-119)}}{2(2)} = \frac{-3 \pm \sqrt{961}}{4}. \] Hence, the solutions are: \[ a = 7, \quad a = -\frac{17}{2}. \]