1. **State the problem:** Find the limit as $n$ approaches infinity of the expression $$\lim_{n \to \infty} \frac{\sqrt{(4n+1)(5+n)(4+n)}}{3n^2 + 2n + 11}.$$\n\n2. **Rewrite the expression:** To analyze the limit, first simplify the expression inside the square root and the denominator.\n\n3. **Analyze the dominant terms:** For large $n$, the dominant terms inside the square root are $4n$, $n$, and $n$, so the product inside the root behaves like $(4n)(n)(n) = 4n^3$.\n\n4. **Approximate the numerator:** $$\sqrt{(4n+1)(5+n)(4+n)} \approx \sqrt{4n^3} = \sqrt{4} \cdot \sqrt{n^3} = 2n^{3/2}.$$\n\n5. **Analyze the denominator:** The dominant term in the denominator is $3n^2$.\n\n6. **Rewrite the limit using dominant terms:** $$\lim_{n \to \infty} \frac{2n^{3/2}}{3n^2} = \lim_{n \to \infty} \frac{2}{3} \cdot \frac{n^{3/2}}{n^2} = \lim_{n \to \infty} \frac{2}{3} \cdot n^{3/2 - 2} = \lim_{n \to \infty} \frac{2}{3} \cdot n^{-1/2}.$$\n\n7. **Simplify the exponent:** Since $n^{-1/2} = \frac{1}{\sqrt{n}}$, as $n \to \infty$, $\frac{1}{\sqrt{n}} \to 0$.\n\n8. **Evaluate the limit:** $$\lim_{n \to \infty} \frac{2}{3} \cdot n^{-1/2} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$