Limits, Continuity & Differentiability
Continuity
Grade None
Question:
<p>Let \(f(x) = \begin{cases} (x-1)^{\frac{1}{2-x}}, & x > 1,\; x \neq 2 \\ k, & x = 2 \end{cases}\). The value of \(k\) for which \(f\) is continuous at \(x = 2\) is</p>
<p>\(1\)</p>
<p>\(e\)</p>
<p>\(e^{-1}\)</p>
<p>\(e^{-2}\)</p>
Step-by-Step Solution
Key Concept: To find the value making f continuous at x=2, evaluate lim(x→2) (x-1)^(1/(2-x)) by rewriting it as e^[(1/(2-x))·ln(x-1)] and analyzing the exponent's behavior as x→2.
<p><strong>Step 1:</strong> For continuity at x = 2, we need k = lim(x→2⁺) (x-1)^(1/(2-x))</p><p><strong>Step 2:</strong> Let y = (x-1)^(1/(2-x)). Take logarithm: ln(y) = [ln(x-1)]/(2-x)</p><p><strong>Step 3:</strong> As x → 2⁺: numerator → ln(1) = 0 and denominator → 0, giving 0/0 form.</p><p><strong>Step 4:</strong> Apply L'Hôpital's rule: lim(x→2⁺) [ln(y)] = lim(x→2⁺) [1/(x-1)]/(-1) = lim(x→2⁺) [-1/(x-1)] = -1</p><p><strong>Step 5:</strong> Therefore lim(x→2⁺) ln(y) = -1, which gives lim(x→2⁺) y = e^(-1) = 1/e</p><p>∴ Answer: k = 1/e (or e⁻¹)</p>
Correct Answer: C