Step 1: Shift the variable:
Let $x = 1 + h$ with $h \to 0$. Then $x - 2 = h - 1$, $x^2 - x = x(x-1) = (1+h)h$, and $x^3 - 3x^2 + 2x = (1+h)h(h-1)$.
Step 2: Combine the two fractions:
The expression is $\frac{h-1}{(1+h)h} - \frac{1}{(1+h)h(h-1)} = \frac{(h-1)^2 - 1}{(1+h)h(h-1)}$.
Step 3: Cancel h and evaluate:
The numerator is $(h-1)^2 - 1 = h^2 - 2h = h(h-2)$. Cancelling $h$ leaves $\frac{h-2}{(1+h)(h-1)}$. As $h \to 0$ this tends to $\frac{-2}{1\cdot(-1)} = 2$.
Step 4: Check the options:
The value is $2$, so option (C) is right. The $0/0$ form comes only from the common factor $h$. After it is removed the function is continuous at $h = 0$, so no other value can appear.
Final Answer:
The limit equals $2$, option (C).
\[ \boxed{2} \]