<p>If \(\lim_{x \to 1} \dfrac{x^2 - ax + b}{x - 1} = 5\), then \(a + b\) is equal to __________.</p>
Step-by-Step Solution
Key Concept: For the limit to exist and be finite as x→1, the numerator must have (x-1) as a factor (making it 0/0 form), which means x=1 must be a root of x²-ax+b. Apply L'Hôpital's rule or factorization to find a and b.
<p><strong>Step 1:</strong> Since the denominator (x-1)→0 as x→1, for the limit to exist and be finite, the numerator must also approach 0.</p><p>Therefore: 1² - a(1) + b = 0 ⟹ <strong>1 - a + b = 0</strong> ... (1)</p><p><strong>Step 2:</strong> Since we have 0/0 form, apply L'Hôpital's rule:</p><p>$$\lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = \lim_{x \to 1} \frac{2x - a}{1} = 2(1) - a = 2 - a$$</p><p><strong>Step 3:</strong> Given that this limit equals 5:</p><p>$$2 - a = 5 ⟹ a = -3$$</p><p><strong>Step 4:</strong> Substitute a = -3 into equation (1):</p><p>$$1 - (-3) + b = 0 ⟹ 1 + 3 + b = 0 ⟹ b = -4$$</p><p><strong>Step 5:</strong> Therefore:</p><p>$$a + b = -3 + (-4) = -7$$</p><p>∴ Answer: <strong>-7</strong></p>
Correct Answer: -7