1. **State the problem:** We want to evaluate the limit $$\lim_{n \to \infty} S_n$$ where $$S_n = \sin\left(\frac{(n+1)\pi}{12n + 11}\right).$$ 2. **Analyze the argument of the sine function:** Consider the argument inside the sine: $$\theta_n = \frac{(n+1)\pi}{12n + 11} = \pi \cdot \frac{n+1}{12n + 11}.$$ 3. **Simplify the fraction inside the argument:** Divide numerator and denominator by $n$: $$\frac{n+1}{12n + 11} = \frac{1 + \frac{1}{n}}{12 + \frac{11}{n}}.$$ 4. **Take the limit of the fraction as $n \to \infty$:** $$\lim_{n \to \infty} \frac{1 + \frac{1}{n}}{12 + \frac{11}{n}} = \frac{1 + 0}{12 + 0} = \frac{1}{12}.$$ 5. **Find the limit of the argument:** $$\lim_{n \to \infty} \theta_n = \pi \cdot \frac{1}{12} = \frac{\pi}{12}.$$ 6. **Evaluate the limit of $S_n$ using continuity of sine:** $$\lim_{n \to \infty} S_n = \sin\left(\lim_{n \to \infty} \theta_n\right) = \sin\left(\frac{\pi}{12}\right).$$ 7. **Simplify $\sin(\pi/12)$:** Recall that $\frac{\pi}{12} = 15^\circ$ and $$\sin\left(\frac{\pi}{12}\right) = \sin(15^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}.$$ **Final answer:** $$\boxed{\lim_{n \to \infty} S_n = \frac{\sqrt{6} - \sqrt{2}}{4}}.$$