Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>Let \( f(x) = \begin{cases} (x-1)\sin\left(\dfrac{1}{x-1}\right), & \text{if } x \neq 1 \\ 0, & \text{if } x = 1 \end{cases} \). Then which one of the following is true?</p>
<p>\(f\) is neither differentiable at \(x = 0\) nor at \(x = 1\).</p>
<p>\(f\) is differentiable at \(x = 0\) and at \(x = 1\).</p>
<p>\(f\) is differentiable at \(x = 0\) but not at \(x = 1\).</p>
<p>\(f\) is differentiable at \(x = 1\) but not at \(x = 0\).</p>
Step-by-Step Solution
Key Concept: Use the squeeze theorem: since $|\sin(\frac{1}{x-1})| \leq 1$, we have $|(x-1)\sin(\frac{1}{x-1})| \leq |x-1|$, which forces the limit to 0 as $x \to 1$. This makes $f$ continuous at $x=1$, and the derivative at $x=1$ can be found using the definition of derivative.
<p><strong>Step 1: Check continuity at x = 1</strong></p><p>We need to verify $\lim_{x \to 1} f(x) = f(1) = 0$.</p><p>Since $|\sin(\frac{1}{x-1})| \leq 1$, we have:</p><p>$$|(x-1)\sin(\frac{1}{x-1})| \leq |x-1| \to 0 \text{ as } x \to 1$$</p><p>By the squeeze theorem: $\lim_{x \to 1} f(x) = 0 = f(1)$</p><p><strong>Step 2: Check differentiability at x = 1</strong></p><p>$$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{h\sin(\frac{1}{h})}{h} = \lim_{h \to 0} \sin(\frac{1}{h})$$</p><p>This limit does not exist (oscillates between -1 and 1).</p><p><strong>Step 3: Conclusion</strong></p><p>$f$ is continuous at $x=1$ but not differentiable at $x=1$.</p><p>∴ Answer: C (continuous at $x=1$ but not differentiable)</p>
Correct Answer: C