Question

\( \displaystyle \lim_{x \to 2} \left( \frac{1}{x-2} - \frac{2}{x^3-3x^2+2x} \right) = \)

• A. \( \frac{2}{3} \)
• B. \( -\frac{2}{3} \)
• C. \( \frac{3}{2} \)
• D. \( -\frac{3}{2} \)

Hint:

In rational-function limits, always factorize completely before substitution. Most indeterminate forms disappear after cancellation of common factors.

Answer 1

The Correct Option is C

Concept: Whenever limits produce forms like \[ \infty-\infty \] we simplify algebraically by factorization and combining fractions. The aim is to remove the factor causing the indeterminate form.

Step 1: Factorizing the cubic polynomial.
Given: \[ x^3-3x^2+2x \] Taking \(x\) common: \[ x(x^2-3x+2) \] Now factorize quadratic: \[ x^2-3x+2=(x-1)(x-2) \] Hence, \[ x^3-3x^2+2x=x(x-1)(x-2) \] Therefore, \[ \lim_{x\to2}\left(\frac1{x-2}-\frac2{x(x-1)(x-2)}\right) \]

Step 2: Taking LCM and simplifying.
Taking common denominator: \[ =\lim_{x\to2}\frac{x(x-1)-2}{x(x-1)(x-2)} \] Expanding numerator: \[ x(x-1)-2=x^2-x-2 \] Factorizing: \[ x^2-x-2=(x-2)(x+1) \] Thus: \[ =\lim_{x\to2}\frac{(x-2)(x+1)}{x(x-1)(x-2)} \] Cancel common factor \((x-2)\): \[ =\lim_{x\to2}\frac{x+1}{x(x-1)} \]

Step 3: Substituting the limit value.
Putting \(x=2\): \[ =\frac{2+1}{2(2-1)} \] \[ =\frac3{2} \] Hence, \[ \boxed{\frac32} \]