Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Given \(f(x) = (x-1)^{\frac{1}{2-x}},\; x > 1,\; x \neq 2\) and \(f(2) = k\). If \(f\) is continuous at \(x = 2\), find \(k\).</p>
<p>\(e\)</p>
<p>\(\dfrac{1}{e}\)</p>
<p>\(e^2\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Rewrite $(x-1)^{\frac{1}{2-x}}$ as $e^{\frac{\ln(x-1)}{2-x}}$ and apply L'Hôpital's rule to find $\lim_{x \to 2^+} \frac{\ln(x-1)}{2-x}$, which has the indeterminate form $\frac{\ln 1}{0}$.
<p><strong>Step 1:</strong> For continuity at $x=2$, we need $\lim_{x \to 2^+} f(x) = f(2) = k$.</p><p><strong>Step 2:</strong> Rewrite the expression: $f(x) = (x-1)^{\frac{1}{2-x}} = e^{\frac{\ln(x-1)}{2-x}}$</p><p><strong>Step 3:</strong> Find $\lim_{x \to 2^+} \frac{\ln(x-1)}{2-x}$. As $x \to 2^+$: numerator $\to \ln(1) = 0$ and denominator $\to 0^-$, giving form $\frac{0}{0}$.</p><p><strong>Step 4:</strong> Apply L'Hôpital's rule: $$\lim_{x \to 2^+} \frac{\ln(x-1)}{2-x} = \lim_{x \to 2^+} \frac{\frac{1}{x-1}}{-1} = \lim_{x \to 2^+} \frac{-1}{x-1} = \frac{-1}{1} = -1$$</p><p><strong>Step 5:</strong> Therefore: $\lim_{x \to 2^+} f(x) = e^{-1} = \frac{1}{e}$</p><p>∴ $k = \frac{1}{e}$ (Answer: B)</p>
Correct Answer: B