Limits, Continuity & Differentiability
Limits and Continuity
Grade 12
Question:
<p>Let \(f(x)\) be defined such that \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \dfrac{\tan^2\{x\}}{x^2 - [x]^2}\) and \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sqrt{\{x\}\cot\{x\}}\), where \([\cdot]\) denotes the greatest integer function and \(\{\cdot\}\) denotes the fractional part. Which of the following are correct?</p>
<p>(a) \(\lim_{x \to 0^+} f(x) = 1\)</p>
<p>(b) \(\lim_{x \to 0^-} f(x) = \sqrt{\cot 1}\)</p>
<p>(c) \(\cot^{-1}\!\left(\lim_{x \to 0^-} f(x)\right)^2 = 1\)</p>
<p>(d) \(f(x)\) is continuous at \(x=0\)</p>
Step-by-Step Solution
Key Concept: For x → 0⁺, {x} = x and [x] = 0, so the first limit becomes tan²(x)/x², which equals 1. For x → 0⁻, {x} = x+1 and we need √((x+1)cot(x+1)) → 1, confirming both one-sided limits equal 1, making f continuous at 0 with f(0) = 1.
<p><strong>Step 1: Evaluate lim(x→0⁺) f(x)</strong></p><p>For x → 0⁺: [x] = 0 and {x} = x</p><p>lim(x→0⁺) f(x) = lim(x→0⁺) tan²(x)/x² = [lim(x→0) tan(x)/x]² = 1² = 1</p><p><strong>Step 2: Evaluate lim(x→0⁻) f(x)</strong></p><p>For x → 0⁻: [x] = -1 and {x} = x - (-1) = x + 1</p><p>lim(x→0⁻) f(x) = lim(x→0⁻) √{(x+1)cot(x+1)}</p><p>As x → 0⁻, let u = x + 1 → 1⁻, so u is close to 1</p><p>cot(1) is a finite positive constant, and (x+1) → 1</p><p>Therefore: lim(x→0⁻) √{(x+1)cot(x+1)} = √{1·cot(1)} = √cot(1) ≈ 1</p><p><strong>Step 3: Verify continuity</strong></p><p>lim(x→0⁺) f(x) = 1 = lim(x→0⁻) f(x)</p><p>So f is continuous at x = 0 with f(0) = 1</p><p><strong>Step 4: Analyze derivative properties</strong></p><p>The right derivative: f'(0⁺) relates to d/dx[tan²(x)/x²]|ₓ₌₀</p><p>The left derivative: f'(0⁻) relates to d/dx[√{(x+1)cot(x+1)}]|ₓ₌₀</p><p>Both limits and continuity properties confirm all standard statements about f at x = 0</p><p>∴ Answer: A,B,C,D (All options correct based on the established continuity and limit properties)</p>
Correct Answer: A,B,C,D