Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>Consider the function \(f(x) = \begin{cases} \max\left\{x, \dfrac{1}{x}\right\}, & \text{when } x \neq 0 \\ \max\left\{x, \dfrac{1}{x}\right\}, & \\ 1, & \text{when } x = 0 \end{cases}\), then:</p>
<p>\(\lim_{x \to 0^+} f(x) \neq 0\)</p>
<p>\(\lim_{x \to 0^-} f(x) = 0\)</p>
<p>\(f(x)\) is continuous for all \(x \neq 0\)</p>
<p>\(f(x)\) is derivable for all \(x \neq 0\)</p>
Step-by-Step Solution
Key Concept: Analyze the function by finding where max{x, 1/x} switches dominance (at x=±1), then check continuity and differentiability at critical points x=0, x=1, x=-1 by examining left/right limits and derivatives.
<p><strong>Step 1: Determine the function form in different regions</strong></p><p>For x > 0: max{x, 1/x} = 1/x when 0 < x < 1; max{x, 1/x} = x when x ≥ 1</p><p>For x < 0: max{x, 1/x} = x when -1 ≤ x < 0; max{x, 1/x} = 1/x when x < -1</p><p><strong>Step 2: Check continuity at x = 0</strong></p><p>lim(x→0⁺) max{x, 1/x} = lim(x→0⁺) 1/x = +∞ ≠ f(0) = 1</p><p>lim(x→0⁻) max{x, 1/x} = lim(x→0⁻) 1/x = -∞ ≠ f(0) = 1</p><p>∴ f is <strong>discontinuous at x = 0</strong></p><p><strong>Step 3: Check continuity and differentiability at x = 1</strong></p><p>lim(x→1⁻) max{x, 1/x} = lim(x→1⁻) 1/x = 1; lim(x→1⁺) max{x, 1/x} = 1</p><p>f(1) = 1, so f is <strong>continuous at x = 1</strong></p><p>For x slightly < 1: f(x) = 1/x, so f'(1⁻) = -1/x²|ₓ₌₁ = -1</p><p>For x slightly > 1: f(x) = x, so f'(1⁺) = 1</p><p>Since f'(1⁻) ≠ f'(1⁺), f is <strong>not differentiable at x = 1</strong></p><p><strong>Step 4: Check continuity and differentiability at x = -1</strong></p><p>By symmetry: f is <strong>continuous but not differentiable at x = -1</strong></p><p>∴ Answer: <strong>A (discontinuous at x=0), C (differentiable everywhere except x=±1,0)</strong></p>
Correct Answer: A,C