Question

\(\underset{x\rightarrow 1}{lim}[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}] =\)

• A. \(3\)
• B. \(4\)
• C. \(2\)
• D. \(1\)

Hint:

Factor both denominators, combine the fractions and cancel the factor (x - 1).

Answer 1

The Correct Option is C

Step 1: Factor the denominators:
\(x^2 - x = x(x-1)\) and \(x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2)\). At \(x = 1\) both fractions are of the form \(\frac{\text{number}}{0}\), so we must combine them before putting \(x = 1\).

Step 2: Take the common denominator:
The common denominator is \(x(x-1)(x-2)\). The first fraction needs a factor \((x-2)\) on top and bottom:
\[ \frac{(x-2)^2 - 1}{x(x-1)(x-2)} \]

Step 3: Simplify the numerator:
\((x-2)^2 - 1 = x^2 - 4x + 3 = (x-1)(x-3)\). So the expression is
\[ \frac{(x-1)(x-3)}{x(x-1)(x-2)} = \frac{x-3}{x(x-2)} \quad (x \ne 1) \]

Step 4: Take the limit:
Now the factor \((x-1)\) is gone, so we can put \(x = 1\):
\[ \frac{1-3}{1\cdot(1-2)} = \frac{-2}{-1} = 2 \]
Options (A), (B) and (D) do not come out of this working. They can only arise from a sign slip, such as reading \(\frac{-2}{-1}\) as \(-2\), or from dropping the \((x-2)\) factor in the numerator.

Final Answer:
The limit is \(2\), which is option (C). \[ \boxed{2} \]