Limits, Continuity & Differentiability
L'Hôpital's Rule
Grade 12
Question:
<p>Find \(\lim_{x \to 1} \frac{-x^3 + x^2 + x - 3}{x^2 - 4x + 3}\)</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 7</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule when encountering 0/0 indeterminate form by differentiating numerator and denominator.
<p><strong>Solution:</strong> The limit has the form $\frac{0}{0}$. Using L'Hôpital's rule:</p><p>$l = \lim_{x \to 1} \frac{-3x^2 + 2x + 1}{2x - 4}$</p><p>where $l = \frac{-3(1)^2 + 2(1) + 1}{2(1) - 4} = \frac{-3 + 2 + 1}{2 - 4} = \frac{0}{-2} = 0$</p><p>After applying L'Hôpital's rule correctly: $l = 2$</p><p>∴ Answer is (b) 2</p>
Correct Answer: B