Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>Given \(f(x) = 5 - |x - 2|\), graph of \(y = f(x)\) is as shown. So, \(f(x)\) is maximum at \(x = 2\), \(\alpha = 2\). Given \(g(x) = |x + 1|\), graph of \(y = g(x)\) is as shown. So, \(g(x)\) is minimum at \(x = -1\), \(\beta = -1\). Therefore, find \(\displaystyle\lim_{x \to -\alpha\beta} \frac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8}\).</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(1\)</p>
<p>\(2\)</p>
Step-by-Step Solution
Key Concept: First identify α and β from the given functions (α = 2, β = -1), then compute -αβ = -2(-1) = 2 to find the limit point. Recognize that the denominator factors to (x-2)(x-4), which vanishes at x = 2, requiring careful cancellation with the numerator.
<p><strong>Step 1: Identify α and β</strong></p><p>From the problem: f(x) = 5 - |x - 2| has maximum at x = 2, so <strong>α = 2</strong></p><p>From the problem: g(x) = |x + 1| has minimum at x = -1, so <strong>β = -1</strong></p><p><strong>Step 2: Calculate the limit point</strong></p><p>-αβ = -(2)(-1) = 2</p><p>So we need: $\lim_{x \to 2} \frac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$</p><p><strong>Step 3: Factor numerator and denominator</strong></p><p>Numerator: $(x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)$</p><p>Denominator: $x^2 - 6x + 8 = (x-2)(x-4)$</p><p><strong>Step 4: Cancel common factor</strong></p><p>$\lim_{x \to 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} = \lim_{x \to 2} \frac{(x-1)(x-3)}{x-4}$</p><p><strong>Step 5: Direct substitution</strong></p><p>$= \frac{(2-1)(2-3)}{2-4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$</p><p>∴ Answer: <strong>B</strong> (which equals 1/2)</p>
Correct Answer: B