To find the limit \(\lim_{x \to 2} \frac{x^3 - 2x^2 + 6x - 12}{x - 2},\) we first substitute \( x = 2 \): \(\frac{2^3 - 2(2^2) + 6(2) - 12}{2 - 2} = \frac{0}{0}.\) This is an indeterminate form, so we factor the numerator. Using synthetic division by \( x - 2 \): \(\begin{array}{r|rrrr} 2 & 1 & -2 & 6 & -12 \\ & & 2 & 0 & 12 \\ \hline & 1 & 0 & 6 & 0 \\ \end{array}\) Thus, \(x^3 - 2x^2 + 6x - 12 = (x - 2)(x^2 + 6).\) We simplify the limit: \(\frac{x^3 - 2x^2 + 6x - 12}{x - 2} = x^2 + 6 \quad \text{for } x \neq 2.\) Now, we find the limit: \(\lim_{x \to 2} (x^2 + 6) = 2^2 + 6 = 10.\) Thus, the limit is 10