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