1. **State the problem:** Find the limit as $x \to 0$ of the expression $$\frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}$$ 2. **Use the conjugate to simplify:** To handle the difference of square roots, multiply numerator and denominator by the conjugates: Numerator conjugate: $\sqrt{1 + x^2} + \sqrt{1 + x}$ Denominator conjugate: $\sqrt{1 + x^3} + \sqrt{1 + x}$ Multiply numerator and denominator by $$\frac{\sqrt{1 + x^2} + \sqrt{1 + x}}{\sqrt{1 + x^3} + \sqrt{1 + x}}$$ 3. **Apply the multiplication:** $$\frac{(\sqrt{1 + x^2} - \sqrt{1 + x})(\sqrt{1 + x^2} + \sqrt{1 + x})}{(\sqrt{1 + x^3} - \sqrt{1 + x})(\sqrt{1 + x^3} + \sqrt{1 + x})} \times \frac{\sqrt{1 + x^3} + \sqrt{1 + x}}{\sqrt{1 + x^2} + \sqrt{1 + x}}$$ Simplify numerator and denominator differences of squares: $$= \frac{(1 + x^2) - (1 + x)}{(1 + x^3) - (1 + x)} \times \frac{\sqrt{1 + x^3} + \sqrt{1 + x}}{\sqrt{1 + x^2} + \sqrt{1 + x}}$$ 4. **Simplify the differences:** $$= \frac{x^2 - x}{x^3 - x} \times \frac{\sqrt{1 + x^3} + \sqrt{1 + x}}{\sqrt{1 + x^2} + \sqrt{1 + x}}$$ 5. **Factor numerator and denominator:** $$x^2 - x = x(x - 1)$$ $$x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)$$ 6. **Cancel common factors:** $$\frac{\cancel{x}(\cancel{x - 1})}{\cancel{x}(\cancel{x - 1})(x + 1)} = \frac{1}{x + 1}$$ So the expression becomes: $$\frac{1}{x + 1} \times \frac{\sqrt{1 + x^3} + \sqrt{1 + x}}{\sqrt{1 + x^2} + \sqrt{1 + x}}$$ 7. **Evaluate the limit as $x \to 0$:** Substitute $x=0$: $$\frac{1}{0 + 1} \times \frac{\sqrt{1 + 0} + \sqrt{1 + 0}}{\sqrt{1 + 0} + \sqrt{1 + 0}} = 1 \times \frac{1 + 1}{1 + 1} = 1 \times 1 = 1$$ **Final answer:** $$\boxed{1}$$