Step 1: Understanding the Concept:
When we plug \( x = 2 \) into the expression, both terms become undefined because their denominators go to zero.
This results in an indeterminate form of the type \( \infty - \infty \).
To evaluate such limits, we must combine the fractions into a single rational expression.
This is done by finding a common denominator, which allows us to simplify the numerator and hopefully cancel out the factor \( (x - 2) \) that is causing the problem.
Step 2: Key Formula or Approach:
Factor the cubic denominator:
\[ x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2) \]
Now we can see that the Least Common Multiple (LCM) of the two denominators is \( x(x - 1)(x - 2) \).
Step 3: Detailed Explanation:
Rewrite the expression with the common denominator:
\[ \lim_{x \to 2} \left[ \frac{x(x - 1)}{x(x - 1)(x - 2)} - \frac{2}{x(x - 1)(x - 2)} \right] \]
\[ = \lim_{x \to 2} \frac{x^2 - x - 2}{x(x - 1)(x - 2)} \]
Now, let us factor the quadratic in the numerator \( x^2 - x - 2 \).
We need two numbers that multiply to -2 and add to -1. These are -2 and 1.
\[ x^2 - x - 2 = (x - 2)(x + 1) \]
Substitute this back into the limit expression:
\[ = \lim_{x \to 2} \frac{(x - 2)(x + 1)}{x(x - 1)(x - 2)} \]
Since \( x \to 2 \), \( x - 2 \neq 0 \), and we can safely cancel the common term \( (x - 2) \):
\[ = \lim_{x \to 2} \frac{x + 1}{x(x - 1)} \]
Now, perform the substitution for \( x = 2 \):
\[ = \frac{2 + 1}{2(2 - 1)} = \frac{3}{2(1)} = \frac{3}{2} \]
Step 4: Final Answer:
The value of the limit is \( \frac{3}{2} \).
This corresponds to option (C).