Limits, Continuity & Differentiability
Continuity and Differentiability at a point
Grade 12
Question:
<p>Let \(f(x) = \dfrac{x}{1+|x|}\). Then \(f(x)\) is</p>
<p>continuous but not differentiable at \(x = 0\)</p>
<p>differentiable at \(x = 0\) but not continuous</p>
<p>continuous and differentiable everywhere</p>
<p>neither continuous nor differentiable at \(x = 0\)</p>
Step-by-Step Solution
Key Concept: Analyze the function separately for x > 0 and x < 0 using the definition of absolute value, then check continuity and differentiability at the critical point x = 0 by computing left and right derivatives.
<p><strong>Step 1:</strong> Express f(x) piecewise using |x|:</p><p>For x ≥ 0: f(x) = x/(1+x)</p><p>For x < 0: f(x) = x/(1−x)</p><p><strong>Step 2:</strong> Check continuity at x = 0:</p><p>lim(x→0⁺) f(x) = 0/(1+0) = 0</p><p>lim(x→0⁻) f(x) = 0/(1−0) = 0</p><p>f(0) = 0, so f is continuous everywhere ✓</p><p><strong>Step 3:</strong> Check differentiability at x = 0:</p><p>Right derivative: f'(0⁺) = d/dx[x/(1+x)]|ₓ₌₀ = [(1+x)−x]/(1+x)²|ₓ₌₀ = 1</p><p>Left derivative: f'(0⁻) = d/dx[x/(1−x)]|ₓ₌₀ = [(1−x)+x]/(1−x)²|ₓ₌₀ = 1</p><p><strong>Step 4:</strong> Since left derivative = right derivative = 1, f is differentiable at x = 0.</p><p>For x ≠ 0, f is clearly differentiable (rational functions).</p><p>∴ Answer: C (f is continuous and differentiable everywhere)</p>
Correct Answer: C