Limits, Continuity & Differentiability
Continuity and differentiability
Grade 12
Question:
<p>Let <i>f</i>, <i>g</i>: <i>R</i> → <i>R</i> be two functions defined by \[f(x) = \begin{cases} x\sin\left(\dfrac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}\] and \(g(x) = xf(x)\).<br><b>Statement 1:</b> <i>f</i> is a continuous function at <i>x</i> = 0.<br><b>Statement 2:</b> <i>g</i> is a differentiable function at <i>x</i> = 0.</p>
<p>Both statements 1 and 2 are false.</p>
<p>Both statements 1 and 2 are true.</p>
<p>Statement 1 is true, statement 2 is false.</p>
<p>Statement 1 is false, statement 2 is true.</p>
Step-by-Step Solution
Key Concept: Use the squeeze theorem for continuity: since |x·sin(1/x)| ≤ |x| → 0 as x→0, f is continuous at 0. For differentiability of g(x) = x²sin(1/x), check if g'(0) exists by examining the limit of [g(x)-g(0)]/x = x·sin(1/x), which equals f(x) and is continuous at 0.
<p><strong>Step 1: Check continuity of f at x = 0</strong></p><p>We need lim(x→0) f(x) = f(0) = 0.</p><p>For x ≠ 0: |f(x)| = |x·sin(1/x)| ≤ |x|·1 = |x|</p><p>By squeeze theorem: lim(x→0) f(x) = 0 ✓</p><p><strong>Statement 1 is TRUE</strong></p><p><strong>Step 2: Check differentiability of g at x = 0</strong></p><p>g(x) = x·f(x) = x²·sin(1/x) for x ≠ 0, and g(0) = 0.</p><p>g'(0) = lim(h→0) [g(h) - g(0)]/h = lim(h→0) [h²sin(1/h)]/h = lim(h→0) h·sin(1/h)</p><p>Since |h·sin(1/h)| ≤ |h| → 0, we have g'(0) = 0 ✓</p><p>For x ≠ 0: g'(x) = 2x·sin(1/x) - cos(1/x), which is continuous at x = 0 because:</p><p>lim(x→0) g'(x) = lim(x→0)[2x·sin(1/x) - cos(1/x)] does not exist (oscillates between -1 and -1 depending on approach)</p><p>However, g'(0) = 0 exists by definition, so g is differentiable at 0.</p><p><strong>Statement 2 is TRUE</strong></p><p>∴ Answer: B (Both statements are true)</p>
Correct Answer: B