Limits, Continuity & Differentiability
Differentiability at a point
Grade 12
<p>Let \(f(x) = x^{\alpha - 1} \cdot \dfrac{e^{2/x} - 1}{e^{2/x} + 1}\) for \(x \neq 0\) and \(f(0) = 0\). For \(f\) to be differentiable at \(x = 0\), the value of \(\alpha\) must satisfy:</p>
Step-by-Step Solution
Key Concept: For differentiability at x=0, f must be continuous at x=0 AND the derivative limit lim[h→0] f(h)/h must exist and be finite. This requires analyzing the behavior of x^(α-1) times a bounded oscillating function near 0.
<p><strong>Step 1: Analyze the oscillating part</strong></p><p>Note that g(x) = (e^(2/x) - 1)/(e^(2/x) + 1) is bounded: -1 < g(x) < 1 for all x ≠ 0.</p><p><strong>Step 2: Check continuity at x=0</strong></p><p>For continuity: lim[x→0] f(x) = lim[x→0] x^(α-1) · g(x) = f(0) = 0</p><p>Since |g(x)| ≤ 1, we need lim[x→0] x^(α-1) = 0, which requires α - 1 > 0, so α > 1.</p><p><strong>Step 3: Check differentiability at x=0</strong></p><p>f'(0) = lim[h→0] f(h)/h = lim[h→0] h^(α-2) · (e^(2/h) - 1)/(e^(2/h) + 1)</p><p>Since the second factor is bounded, we need h^(α-2) → 0 as h→0, which requires α - 2 > 0, so α > 2.</p><p><strong>Step 4: Verify sufficiency</strong></p><p>When α > 2: both f(x)→0 (continuity) and f'(0) = 0 exist (differentiability). For α ≤ 2, either continuity fails (α ≤ 1) or the derivative doesn't exist (1 < α ≤ 2).</p><p>∴ Answer: <strong>α > 2</strong> (or α ∈ (2, ∞))</p>
Correct Answer: B