Limits, Continuity & Differentiability
Differentiability of functions
Grade 12
Question:
<p>If \(f(x) = (x-1)\sin\left(\dfrac{1}{x-1}\right)\) when \(x \neq 1\) and \(f(1) = 0\), then which of the following is true?</p>
<p>\(f\) is differentiable at \(x=1\) but not at \(x=0\)</p>
<p>\(f\) is not differentiable at \(x=1\)</p>
<p>\(f\) is differentiable at both \(x=0\) and \(x=1\)</p>
<p>\(f\) is differentiable at \(x=0\) but not at \(x=1\)</p>
Step-by-Step Solution
Key Concept: A function is continuous at a point if the limit equals the function value there; use the squeeze theorem since |sin(u)| ≤ 1 for any u.
<p><strong>Step 1:</strong> Check continuity at x = 1 by evaluating lim(x→1) f(x).</p><p><strong>Step 2:</strong> For x ≠ 1: f(x) = (x-1)sin(1/(x-1)). Since |sin(1/(x-1))| ≤ 1 for all x ≠ 1, we have:</p><p>|f(x)| = |x-1| · |sin(1/(x-1))| ≤ |x-1|</p><p><strong>Step 3:</strong> By squeeze theorem: as x → 1, |f(x)| → 0, so lim(x→1) f(x) = 0 = f(1).</p><p><strong>Step 4:</strong> Therefore f is continuous at x = 1. However, differentiability requires lim(x→1) [f(x)-f(1)]/(x-1) = lim(x→1) sin(1/(x-1)) to exist. This limit does NOT exist since sin(1/(x-1)) oscillates as x → 1, so f is NOT differentiable at x = 1.</p><p>∴ Answer: B (f is continuous but not differentiable at x = 1)</p>
Correct Answer: B