Limits, Continuity & Differentiability
Continuity and Differentiability at a point
Grade 12
Question:
<p>Let <br>
\[ f(x) = \begin{cases} xe^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases} \]
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 becomes -2/x → -∞, so e^(-2/x) → 0 faster than any polynomial. For x < 0: the exponent becomes 0, so e^0 = 1, giving f(x) = x. This asymmetry determines differentiability at x = 0.
<p><strong>Step 1: Simplify f(x) by cases using |x|/x</strong></p><p>For x > 0: |x| = x, so |x|/x = 1, thus exponent = -(1/x + 1/x) = -2/x</p><p>For x < 0: |x| = -x, so |x|/x = -1, thus exponent = -(-1/x + 1/x) = 0</p><p>Therefore: f(x) = {xe^(-2/x) for x > 0; x for x < 0; 0 for x = 0}</p><p><strong>Step 2: Check continuity at x = 0</strong></p><p>lim(x→0+) xe^(-2/x) = 0 (exponential decay dominates linear growth)</p><p>lim(x→0-) x = 0</p><p>f(0) = 0 ✓ Continuous at x = 0</p><p><strong>Step 3: Check differentiability at x = 0</strong></p><p>Right derivative: lim(h→0+) [he^(-2/h) - 0]/h = lim(h→0+) e^(-2/h) = 0</p><p>Left derivative: lim(h→0-) [h - 0]/h = lim(h→0-) 1 = 1</p><p>Left and right derivatives are unequal (1 ≠ 0) ∴ f is not differentiable at x = 0</p><p><strong>Step 4: Check differentiability for x ≠ 0</strong></p><p>For x > 0: f'(x) = e^(-2/x) + x·e^(-2/x)·(2/x²) = e^(-2/x)(1 + 2/x) — exists</p><p>For x < 0: f'(x) = 1 — exists</p><p>∴ Answer: A (f is continuous everywhere but not differentiable at x = 0)</p>
Correct Answer: A