Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \( f(x) = \begin{cases} xe^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases} \) then \( f(x) \) is</p>
<p>continuous as well as differentiable for all \(x\).</p>
<p>continuous for all \(x\) but not differentiable at \(x = 0\).</p>
<p>neither differentiable nor continuous at \(x = 0\).</p>
<p>discontinuous everywhere.</p>
Step-by-Step Solution
Key Concept: For x < 0: the exponent -1/|x| - 1/x = -1/(-x) - 1/x = 1/x - 1/x = 0, making f(x) = x·e⁰ = x. This linear behavior near 0 is key to checking differentiability.
<p><strong>Step 1: Analyze f(x) for x ≠ 0</strong></p><p>For <strong>x > 0</strong>: f(x) = x·e^(-1/x - 1/x) = x·e^(-2/x) → 0 as x → 0⁺</p><p>For <strong>x < 0</strong>: |x| = -x, so -1/|x| - 1/x = -1/(-x) - 1/x = 1/x - 1/x = 0<br/>Therefore f(x) = x·e⁰ = <strong>x</strong></p><p><strong>Step 2: Check continuity at x = 0</strong></p><p>lim(x→0⁻) f(x) = lim(x→0⁻) x = 0 = f(0) ✓<br/>lim(x→0⁺) f(x) = lim(x→0⁺) x·e^(-2/x) = 0·0 = 0 = f(0) ✓<br/><strong>f is continuous at x = 0</strong></p><p><strong>Step 3: Check differentiability at x = 0</strong></p><p><strong>Left derivative:</strong> f'(0⁻) = lim(h→0⁻) [h - 0]/h = 1</p><p><strong>Right derivative:</strong> f'(0⁺) = lim(h→0⁺) [h·e^(-2/h) - 0]/h = lim(h→0⁺) e^(-2/h) = 0<br/>(since e^(-2/h) → 0 faster than any polynomial as h → 0⁺)</p><p>Since f'(0⁻) = 1 ≠ 0 = f'(0⁺), <strong>f is not differentiable at x = 0</strong></p><p><strong>Conclusion:</strong> f(x) is <strong>continuous but not differentiable at x = 0</strong><br/>∴ Answer: A</p>
Correct Answer: A