2. Evaluate each of the following limits (a) \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2} \) (b) \( \lim _{x \rightarrow 0} \frac{(x+4)^{2}-16}{x} \) (c) \( \lim _{x \rightarrow 3} \frac{x^{3}-9 x}{3 x^{2}-6 x-9} \) 3. Evaluate each of the following limits (a) \( \lim _{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x} \) (b) \( \lim _{x \rightarrow 2} \frac{\sqrt{x+2}-2}{x-2} \) (c) \( \lim _{x \rightarrow 1} \frac{\sqrt{x+3}-2}{x-1} \)

**2. (a) Evaluate \(\displaystyle \lim_{x \rightarrow 2} \frac{x^{2}-4}{x-2}\)** Factor the numerator as a difference of squares: \[ x^2 - 4 = (x-2)(x+2) \] Then \[ \frac{x^{2}-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (x \neq 2) \] Taking the limit as \(x \rightarrow 2\), \[ \lim_{x \rightarrow 2} (x+2) = 2+2 = 4. \] --- **2. (b) Evaluate \(\displaystyle \lim_{x \rightarrow 0} \frac{(x+4)^{2}-16}{x}\)** Expand the numerator: \[ (x+4)^2 - 16 = (x^2 + 8x + 16) - 16 = x^2 + 8x. \] Factor out \(x\): \[ x^2 + 8x = x(x+8). \] Thus, \[ \frac{(x+4)^{2}-16}{x} = \frac{x(x+8)}{x} = x+8 \quad (x \neq 0). \] Taking the limit as \(x \rightarrow 0\), \[ \lim_{x \rightarrow 0} (x+8) = 0+8 = 8. \] --- **2. (c) Evaluate \(\displaystyle \lim_{x \rightarrow 3} \frac{x^{3}-9 x}{3 x^{2}-6 x-9}\)** Factor the numerator: \[ x^3 - 9x = x(x^2 - 9) = x(x-3)(x+3). \] Factor the denominator. First, factor out \(3\): \[ 3x^2 - 6x - 9 = 3(x^2 - 2x - 3). \] Factor the quadratic: \[ x^2 - 2x - 3 = (x-3)(x+1). \] Thus, the denominator becomes: \[ 3(x-3)(x+1). \] Cancelling the common factor \((x-3)\) (valid for \(x \neq 3\)): \[ \frac{x(x-3)(x+3)}{3(x-3)(x+1)} = \frac{x(x+3)}{3(x+1)}. \] Taking the limit as \(x \rightarrow 3\): \[ \lim_{x \rightarrow 3} \frac{x(x+3)}{3(x+1)} = \frac{3(3+3)}{3(3+1)} = \frac{3 \times 6}{3 \times 4} = \frac{18}{12} = \frac{3}{2}. \] --- **3. (a) Evaluate \(\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}\)** Multiply numerator and denominator by the conjugate \(\sqrt{x+1}+1\): \[ \frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \frac{(x+1)-1}{x(\sqrt{x+1}+1)} = \frac{x}{x(\sqrt{x+1}+1)} = \frac{1}{\sqrt{x+1}+1}. \] Taking the limit as \(x \rightarrow 0\): \[ \lim_{x \rightarrow 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{0+1}+1} = \frac{1}{1+1} = \frac{1}{2}. \] --- **3. (b) Evaluate \(\displaystyle \lim_{x \rightarrow 2} \frac{\sqrt{x+2}-2}{x-2}\)** Multiply the expression by the conjugate of the numerator: \[ \frac{\sqrt{x+2}-2}{x-2} \cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2} = \frac{(x+2)-4}{(x-2)(\sqrt{x+2}+2)} = \frac{x-2}{(x-2)(\sqrt{x+2}+2)} = \frac{1}{\sqrt{x+2}+2}. \] Taking the limit as \(x \rightarrow 2\): \[ \lim_{x \rightarrow 2} \frac{1}{\sqrt{x+2}+2} = \frac{1}{\sqrt{2+2}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}. \] --- **3. (c) Evaluate \(\displaystyle \lim_{x \rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}\)** Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{x+3}-2}{x-1} \cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} = \frac{(x+3)-4}{(x-1)(\sqrt{x+3}+2)} = \frac{x-1}{(x-1)(\sqrt{x+3}+2)} = \frac{1}{\sqrt{x+3}+2}. \] Taking the limit as \(x \rightarrow 1\): \[ \lim_{x \rightarrow 1} \frac{1}{\sqrt{x+3}+2} = \frac{1}{\sqrt{1+3}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}. \]