1. **State the problem:** We want to find the limit as $n$ approaches infinity of the expression $$\frac{n^8 + n^2 + 1}{n^6 + 2} \cdot \frac{2n + 3}{n^3 + 1}.$$\n\n2. **Recall the rule for limits of rational functions:** When $n \to \infty$, the highest power terms dominate the behavior of polynomials. We focus on the leading terms in numerator and denominator.\n\n3. **Analyze each fraction separately:**\n- For the first fraction: $$\frac{n^8 + n^2 + 1}{n^6 + 2} \approx \frac{n^8}{n^6} = n^2.$$\n- For the second fraction: $$\frac{2n + 3}{n^3 + 1} \approx \frac{2n}{n^3} = \frac{2}{n^2}.$$\n\n4. **Multiply the approximations:** $$n^2 \cdot \frac{2}{n^2} = 2.$$\n\n5. **More rigorous approach:** Divide numerator and denominator of each fraction by the highest power of $n$ in the denominator to confirm:\n\n$$\frac{n^8 + n^2 + 1}{n^6 + 2} = \frac{n^6(n^2 + n^{-4} + n^{-6})}{n^6(1 + 2n^{-6})} = \frac{n^2 + n^{-4} + n^{-6}}{1 + 2n^{-6}} \to n^2$$ as $n \to \infty$.\n\n$$\frac{2n + 3}{n^3 + 1} = \frac{n^3(2n^{-2} + 3n^{-3})}{n^3(1 + n^{-3})} = \frac{2n^{-2} + 3n^{-3}}{1 + n^{-3}} \to 0$$ but since we multiply by $n^2$ from the first fraction, the product tends to a finite limit.\n\n6. **Combine the exact expressions:**\n$$\lim_{n \to \infty} \frac{n^8 + n^2 + 1}{n^6 + 2} \cdot \frac{2n + 3}{n^3 + 1} = \lim_{n \to \infty} \left(\frac{n^2 + n^{-4} + n^{-6}}{1 + 2n^{-6}} \cdot \frac{2n^{-2} + 3n^{-3}}{1 + n^{-3}}\right).$$\n\n7. **Simplify the product inside the limit:**\n$$= \lim_{n \to \infty} \frac{(n^2 + n^{-4} + n^{-6})(2n^{-2} + 3n^{-3})}{(1 + 2n^{-6})(1 + n^{-3})}.$$\n\n8. **Multiply numerator terms:**\n$$ (n^2)(2n^{-2}) = 2, \quad (n^2)(3n^{-3}) = 3n^{-1}, \quad n^{-4}(2n^{-2}) = 2n^{-6}, \quad n^{-4}(3n^{-3}) = 3n^{-7}, \quad n^{-6}(2n^{-2}) = 2n^{-8}, \quad n^{-6}(3n^{-3}) = 3n^{-9}.$$\n\n9. **Sum numerator terms:**\n$$2 + 3n^{-1} + 2n^{-6} + 3n^{-7} + 2n^{-8} + 3n^{-9}.$$\n\n10. **As $n \to \infty$, all terms with negative powers vanish, so numerator $\to 2$. Denominator $\to 1 \cdot 1 = 1$.**\n\n11. **Final limit:** $$\boxed{2}.$$