1. **State the problems:** We need to find the limits: $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x}$$ and $$\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}$$ 2. **Recall the conjugate multiplication technique:** When limits involve expressions like $\sqrt{a} - b$, multiplying numerator and denominator by the conjugate $\sqrt{a} + b$ helps simplify. --- ### First limit: 3. Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x+1}-1}{x} \times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \frac{(\sqrt{x+1})^2 - 1^2}{x(\sqrt{x+1}+1)} = \frac{x+1 - 1}{x(\sqrt{x+1}+1)} = \frac{x}{x(\sqrt{x+1}+1)}$$ 4. Cancel $x$ in numerator and denominator: $$\frac{\cancel{x}}{\cancel{x}(\sqrt{x+1}+1)} = \frac{1}{\sqrt{x+1}+1}$$ 5. Evaluate the limit as $x \to 0$: $$\lim_{x\to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{0+1}+1} = \frac{1}{1+1} = \frac{1}{2}$$ --- ### Second limit: 6. Multiply numerator and denominator by the conjugate of the denominator: $$\frac{x-4}{\sqrt{x}-2} \times \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)(\sqrt{x}+2)}{(\sqrt{x})^2 - 2^2} = \frac{(x-4)(\sqrt{x}+2)}{x - 4}$$ 7. Cancel $x-4$ in numerator and denominator: $$\frac{\cancel{x-4}(\sqrt{x}+2)}{\cancel{x-4}} = \sqrt{x} + 2$$ 8. Evaluate the limit as $x \to 4$: $$\lim_{x\to 4} (\sqrt{x} + 2) = \sqrt{4} + 2 = 2 + 2 = 4$$