Step 1: Factorizing the numerator and denominator \( x^3 - 27 = (x-3)(x^2 + 3x + 9). \) \( x^2 - 9 = (x-3)(x+3). \)
Step 2: Cancel common terms \( \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}. \) For \( x \neq 3 \), canceling \( (x-3) \), \( \lim_{x \to 3} \frac{x^2+3x+9}{x+3}. \)
Step 3: Substitute \( x = 3 \) \( \frac{3^2+3(3)+9}{3+3} = \frac{9+9+9}{6} = \frac{27}{6} = \frac{9}{2}. \)