1. **Problem statement:** Find the limit as $x \to 0$ of $$\frac{\sqrt{x^2 - 3x + 1} - \sqrt{x^2 + 5x + 1}}{2x}.$$ 2. **Formula and approach:** When limits involve differences of square roots, multiply numerator and denominator by the conjugate to simplify. The conjugate here is $$\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}.$$ 3. **Multiply numerator and denominator by the conjugate:** $$\frac{\sqrt{x^2 - 3x + 1} - \sqrt{x^2 + 5x + 1}}{2x} \times \frac{\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}}{\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}} = \frac{(x^2 - 3x + 1) - (x^2 + 5x + 1)}{2x \left(\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}\right)}.$$ 4. **Simplify numerator:** $$x^2 - 3x + 1 - x^2 - 5x - 1 = -8x.$$ 5. **Rewrite the expression:** $$\frac{-8x}{2x \left(\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}\right)}.$$ 6. **Cancel common factor $x$:** $$\frac{\cancel{-8x}}{2\cancel{x} \left(\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}\right)} = \frac{-8}{2 \left(\sqrt{x^2 - 3x + 1} + \sqrt{x^2 + 5x + 1}\right)}.$$ 7. **Evaluate the limit as $x \to 0$ by substituting $x=0$ inside the square roots:** $$\sqrt{0 - 0 + 1} + \sqrt{0 + 0 + 1} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2.$$ 8. **Final limit value:** $$\lim_{x \to 0} \frac{\sqrt{x^2 - 3x + 1} - \sqrt{x^2 + 5x + 1}}{2x} = \frac{-8}{2 \times 2} = \frac{-8}{4} = -2.$$ --- 9. **Problem statement:** Find the limit as $x \to \infty$ of $$\sqrt{x^2 + 6x + 2} - (x - 5).$$ 10. **Approach:** For limits at infinity involving square roots, factor out $x^2$ inside the root to simplify. 11. **Rewrite the square root:** $$\sqrt{x^2 + 6x + 2} = \sqrt{x^2 \left(1 + \frac{6}{x} + \frac{2}{x^2}\right)} = |x| \sqrt{1 + \frac{6}{x} + \frac{2}{x^2}}.$$ 12. **Since $x \to \infty$, $|x| = x$, so:** $$\sqrt{x^2 + 6x + 2} = x \sqrt{1 + \frac{6}{x} + \frac{2}{x^2}}.$$ 13. **Use binomial expansion for the square root for large $x$:** $$\sqrt{1 + u} \approx 1 + \frac{u}{2} - \frac{u^2}{8} + \cdots$$ where $$u = \frac{6}{x} + \frac{2}{x^2}.$$ 14. **Approximate:** $$\sqrt{1 + \frac{6}{x} + \frac{2}{x^2}} \approx 1 + \frac{1}{2} \left(\frac{6}{x} + \frac{2}{x^2}\right) = 1 + \frac{3}{x} + \frac{1}{x^2}.$$ 15. **Multiply by $x$:** $$x \left(1 + \frac{3}{x} + \frac{1}{x^2}\right) = x + 3 + \frac{1}{x}.$$ 16. **Rewrite the original expression:** $$\sqrt{x^2 + 6x + 2} - (x - 5) \approx (x + 3 + \frac{1}{x}) - (x - 5) = 3 + \frac{1}{x} + 5 = 8 + \frac{1}{x}.$$ 17. **Take the limit as $x \to \infty$:** $$\lim_{x \to \infty} 8 + \frac{1}{x} = 8 + 0 = 8.$$ --- **Final answers:** - For problem 4: $$\boxed{-2}.$$ - For problem 5: $$\boxed{8}.$$