Limit Evaluation 91Cb9B

1. **Problem:** Evaluate the limit $$\lim_{x \to 4} \left( \frac{2x^2 - 4x - 24}{x^2 - 16} - \frac{1}{4 - x} \right)$$ 2. **Formula and rules:** To evaluate limits involving rational expressions, factor polynomials and simplify expressions. Remember that $$x^2 - 16 = (x - 4)(x + 4)$$ and that $$4 - x = -(x - 4)$$. 3. **Work:** \[ \frac{2x^2 - 4x - 24}{x^2 - 16} - \frac{1}{4 - x} = \frac{2(x^2 - 2x - 12)}{(x - 4)(x + 4)} - \frac{1}{4 - x} \] Factor numerator: \[ x^2 - 2x - 12 = (x - 4)(x + 3) \] So: \[ \frac{2(x - 4)(x + 3)}{(x - 4)(x + 4)} - \frac{1}{4 - x} \] Cancel $$x - 4$$: \[ \frac{\cancel{2(x - 4)}(x + 3)}{\cancel{(x - 4)}(x + 4)} - \frac{1}{4 - x} = \frac{2(x + 3)}{x + 4} - \frac{1}{4 - x} \] Rewrite $$\frac{1}{4 - x} = -\frac{1}{x - 4}$$: \[ \frac{2(x + 3)}{x + 4} + \frac{1}{x - 4} \] Find common denominator $$(x + 4)(x - 4)$$: \[ \frac{2(x + 3)(x - 4)}{(x + 4)(x - 4)} + \frac{x + 4}{(x - 4)(x + 4)} = \frac{2(x + 3)(x - 4) + (x + 4)}{(x + 4)(x - 4)} \] Expand numerator: \[ 2(x + 3)(x - 4) + (x + 4) = 2(x^2 - 4x + 3x - 12) + x + 4 = 2(x^2 - x - 12) + x + 4 = 2x^2 - 2x - 24 + x + 4 = 2x^2 - x - 20 \] So the expression is: \[ \frac{2x^2 - x - 20}{(x + 4)(x - 4)} \] Evaluate limit as $$x \to 4$$: \[ \frac{2(4)^2 - 4 - 20}{(4 + 4)(4 - 4)} = \frac{2(16) - 4 - 20}{8 \times 0} = \frac{32 - 4 - 20}{0} = \frac{8}{0} \] Since denominator tends to zero and numerator tends to 8 (nonzero), the limit tends to infinity or does not exist. 4. **Conclusion:** The limit does not exist (infinite). --- 1. **Problem:** Evaluate $$\lim_{x \to 0} \frac{7x}{\sqrt{3x + 4} - 2}$$ 2. **Formula and rules:** Use conjugate multiplication to simplify limits involving square roots. 3. **Work:** Multiply numerator and denominator by conjugate: \[ \frac{7x}{\sqrt{3x + 4} - 2} \times \frac{\sqrt{3x + 4} + 2}{\sqrt{3x + 4} + 2} = \frac{7x(\sqrt{3x + 4} + 2)}{(\sqrt{3x + 4} - 2)(\sqrt{3x + 4} + 2)} \] Denominator simplifies: \[ (\sqrt{3x + 4})^2 - 2^2 = 3x + 4 - 4 = 3x \] So expression is: \[ \frac{7x(\sqrt{3x + 4} + 2)}{3x} \] Cancel $$x$$: \[ \frac{\cancel{7x}(\sqrt{3x + 4} + 2)}{3\cancel{x}} = \frac{7}{3}(\sqrt{3x + 4} + 2) \] Evaluate limit as $$x \to 0$$: \[ \frac{7}{3}(\sqrt{4} + 2) = \frac{7}{3}(2 + 2) = \frac{7}{3} \times 4 = \frac{28}{3} \] 4. **Conclusion:** The limit is $$\frac{28}{3}$$. --- 1. **Problem:** Evaluate $$\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$$ 2. **Formula and rules:** This is the definition of the derivative of $$\sqrt{x}$$. Use conjugate multiplication. 3. **Work:** Multiply numerator and denominator by conjugate: \[ \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} = \frac{h}{h(\sqrt{x + h} + \sqrt{x})} \] Cancel $$h$$: \[ \frac{1}{\sqrt{x + h} + \sqrt{x}} \] Evaluate limit as $$h \to 0$$: \[ \frac{1}{2\sqrt{x}} \] 4. **Conclusion:** The limit is $$\frac{1}{2\sqrt{x}}$$. --- 1. **Problem:** Evaluate $$\lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{3 - x^2}}{x - 1}$$ 2. **Formula and rules:** Use conjugate multiplication and recognize this as a derivative form. 3. **Work:** Multiply numerator and denominator by conjugate: \[ \frac{\sqrt{2x} - \sqrt{3 - x^2}}{x - 1} \times \frac{\sqrt{2x} + \sqrt{3 - x^2}}{\sqrt{2x} + \sqrt{3 - x^2}} = \frac{2x - (3 - x^2)}{(x - 1)(\sqrt{2x} + \sqrt{3 - x^2})} = \frac{2x - 3 + x^2}{(x - 1)(\sqrt{2x} + \sqrt{3 - x^2})} \] Factor numerator: \[ x^2 + 2x - 3 = (x + 3)(x - 1) \] Cancel $$x - 1$$: \[ \frac{(x + 3)\cancel{(x - 1)}}{\cancel{(x - 1)}(\sqrt{2x} + \sqrt{3 - x^2})} = \frac{x + 3}{\sqrt{2x} + \sqrt{3 - x^2}} \] Evaluate limit as $$x \to 1$$: \[ \frac{1 + 3}{\sqrt{2(1)} + \sqrt{3 - 1}} = \frac{4}{\sqrt{2} + \sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] 4. **Conclusion:** The limit is $$\sqrt{2}$$. --- 1. **Problem:** Evaluate $$\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{2x^2}$$ 2. **Formula and rules:** Use conjugate multiplication to simplify. 3. **Work:** Multiply numerator and denominator by conjugate: \[ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{2x^2} \times \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} = \frac{(1 + x^2) - (1 - x^2)}{2x^2(\sqrt{1 + x^2} + \sqrt{1 - x^2})} = \frac{2x^2}{2x^2(\sqrt{1 + x^2} + \sqrt{1 - x^2})} \] Cancel $$2x^2$$: \[ \frac{1}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \] Evaluate limit as $$x \to 0$$: \[ \frac{1}{\sqrt{1} + \sqrt{1}} = \frac{1}{2} \] 4. **Conclusion:** The limit is $$\frac{1}{2}$$. --- 1. **Problem:** Evaluate $$\lim_{x \to 2} \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4}$$ 2. **Formula and rules:** Use conjugate multiplication to simplify numerator and denominator. 3. **Work:** Multiply numerator and denominator by conjugate of denominator: \[ \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4} \times \frac{\sqrt{x^2 + 12} + 4}{\sqrt{x^2 + 12} + 4} = \frac{(x - \sqrt{8 - x^2})(\sqrt{x^2 + 12} + 4)}{(\sqrt{x^2 + 12})^2 - 4^2} = \frac{(x - \sqrt{8 - x^2})(\sqrt{x^2 + 12} + 4)}{x^2 + 12 - 16} = \frac{(x - \sqrt{8 - x^2})(\sqrt{x^2 + 12} + 4)}{x^2 - 4} \] Factor denominator: \[ x^2 - 4 = (x - 2)(x + 2) \] Multiply numerator and denominator by conjugate of numerator: \[ (x - \sqrt{8 - x^2}) \times (x + \sqrt{8 - x^2}) = x^2 - (8 - x^2) = 2x^2 - 8 \] Rewrite original expression as: \[ \frac{(x - \sqrt{8 - x^2})(\sqrt{x^2 + 12} + 4)}{(x - 2)(x + 2)} = \frac{2x^2 - 8}{(x + \sqrt{8 - x^2})(x - 2)(x + 2)(\sqrt{x^2 + 12} + 4)} \] Factor numerator: \[ 2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2) \] Cancel $$ (x - 2)(x + 2) $$: \[ \frac{2\cancel{(x - 2)}\cancel{(x + 2)}}{(x + \sqrt{8 - x^2})\cancel{(x - 2)}\cancel{(x + 2)}(\sqrt{x^2 + 12} + 4)} = \frac{2}{(x + \sqrt{8 - x^2})(\sqrt{x^2 + 12} + 4)} \] Evaluate limit as $$x \to 2$$: \[ \frac{2}{(2 + \sqrt{8 - 4})(\sqrt{4 + 12} + 4)} = \frac{2}{(2 + \sqrt{4})(\sqrt{16} + 4)} = \frac{2}{(2 + 2)(4 + 4)} = \frac{2}{4 \times 8} = \frac{2}{32} = \frac{1}{16} \] 4. **Conclusion:** The limit is $$\frac{1}{16}$$.