1. **State the problem:** Find the limit as $x \to 0$ of the expression $$\frac{\sqrt{1+x^2} - \sqrt{1+x}}{\sqrt{1+x^3} - \sqrt{1+x}}.$$\n\n2. **Recall the conjugate multiplication technique:** To simplify expressions with differences of square roots, multiply numerator and denominator by the conjugate of numerator and denominator respectively to rationalize.\n\n3. **Multiply numerator and denominator by the conjugates:**\n$$\frac{\sqrt{1+x^2} - \sqrt{1+x}}{\sqrt{1+x^3} - \sqrt{1+x}} \times \frac{\sqrt{1+x^2} + \sqrt{1+x}}{\sqrt{1+x^2} + \sqrt{1+x}} \times \frac{\sqrt{1+x^3} + \sqrt{1+x}}{\sqrt{1+x^3} + \sqrt{1+x}}.$$\n\n4. **Simplify numerator:**\n$$\left(\sqrt{1+x^2} - \sqrt{1+x}\right)\left(\sqrt{1+x^2} + \sqrt{1+x}\right) = (1+x^2) - (1+x) = x^2 - x.$$\n\n5. **Simplify denominator:**\n$$\left(\sqrt{1+x^3} - \sqrt{1+x}\right)\left(\sqrt{1+x^3} + \sqrt{1+x}\right) = (1+x^3) - (1+x) = x^3 - x.$$\n\n6. **Rewrite the expression:**\n$$\frac{x^2 - x}{x^3 - x} \times \frac{\sqrt{1+x^3} + \sqrt{1+x}}{\sqrt{1+x^2} + \sqrt{1+x}}.$$\n\n7. **Factor numerator and denominator:**\n$$x^2 - x = x(x-1), \quad x^3 - x = x(x^2 - 1) = x(x-1)(x+1).$$\n\n8. **Cancel common factors:**\n$$\frac{\cancel{x}(\cancel{x-1})}{\cancel{x}(\cancel{x-1})(x+1)} = \frac{1}{x+1}.$$\n\n9. **Substitute back:**\n$$\frac{1}{x+1} \times \frac{\sqrt{1+x^3} + \sqrt{1+x}}{\sqrt{1+x^2} + \sqrt{1+x}}.$$\n\n10. **Evaluate the limit as $x \to 0$:**\n$$\lim_{x \to 0} \frac{1}{x+1} = 1,$$\n$$\lim_{x \to 0} \sqrt{1+x^3} + \sqrt{1+x} = \sqrt{1+0} + \sqrt{1+0} = 1 + 1 = 2,$$\n$$\lim_{x \to 0} \sqrt{1+x^2} + \sqrt{1+x} = \sqrt{1+0} + \sqrt{1+0} = 1 + 1 = 2.$$\n\n11. **Final limit:**\n$$1 \times \frac{2}{2} = 1.$$\n\n**Answer:** The limit is $1$.