Question

The value of \[ \lim_{x \to 1} \frac{x^{3} - 3x + 2}{x^{3} - x^{2} - x + 1} \] is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

If both numerator and denominator repeat $0/0$ after the first differentiation, apply L'Hôpital's rule again until the limit becomes finite.

Answer 1

Correct Answer: 1.4 - 1.6

Substitute $x = 1$: \[ \text{Numerator} = 1^{3} - 3(1) + 2 = 0, \text{Denominator} = 1^{3} - 1^{2} - 1 + 1 = 0. \] This is a $0/0$ indeterminate form → apply L'Hôpital's rule. Differentiate numerator: \[ \frac{d}{dx}(x^{3} - 3x + 2) = 3x^{2} - 3. \] Differentiate denominator: \[ \frac{d}{dx}(x^{3} - x^{2} - x + 1) = 3x^{2} - 2x - 1. \] Now substitute $x = 1$: \[ \frac{3(1)^{2} - 3}{3(1)^{2} - 2(1) - 1} = \frac{0}{0} = \text{indeterminate}. \] Apply L'Hôpital's rule again. Second derivatives: \[ \frac{d}{dx}(3x^{2} - 3) = 6x, \frac{d}{dx}(3x^{2} - 2x - 1) = 6x - 2. \] Evaluate at $x=1$: \[ \frac{6}{6 - 2} = \frac{6}{4} = 1.5. \] Rounded to one decimal place: \[ 1.5. \]