Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p><strong>279.</strong> Let \( f(x) = \dfrac{x \ln x - \ln x}{9x^2 - 2e^x x - 9x + 2e^x} + 2 \) and \( g(x) = \sin^2\left(\dfrac{\pi x^2}{2}\right) \), then the value of \( \lim_{x \to 1} \dfrac{f(x)}{g(x)} \) is:</p>
<p>(a) 2</p>
<p>(b) \(\dfrac{1}{3}\)</p>
<p>(c) 3</p>
<p>(d) \(\dfrac{2}{3}\)</p>
Step-by-Step Solution
Key Concept: Factor the numerator as ln(x)(x-1) and denominator as (x-1)(9x-2e^x), then cancel (x-1) to resolve the 0/0 indeterminate form without requiring L'Hôpital's rule.
<p><strong>Step 1: Simplify f(x) - 2</strong></p><p>Numerator: x ln(x) - ln(x) = ln(x)(x - 1)</p><p><strong>Step 2: Factor the denominator</strong></p><p>9x² - 2e^x·x - 9x + 2e^x = 9x(x-1) - 2e^x(x-1) = (x-1)(9x - 2e^x)</p><p><strong>Step 3: Simplify the fraction</strong></p><p>f(x) - 2 = ln(x)(x-1)/[(x-1)(9x - 2e^x)] = ln(x)/(9x - 2e^x)</p><p>Therefore: f(x) = ln(x)/(9x - 2e^x) + 2</p><p><strong>Step 4: Evaluate limits at x = 1</strong></p><p>f(1) = ln(1)/(9 - 2e) + 2 = 0 + 2 = 2</p><p>g(1) = sin²(π/2) = 1</p><p><strong>Step 5: Calculate the limit</strong></p><p>lim(x→1) f(x)/g(x) = f(1)/g(1) = 2/1 = 2</p><p>∴ Answer: B</p>
Correct Answer: B