<p>If \(\lim_{x \to 2} \dfrac{\tan(x-2)\{x^2 + (k-2)x - 2k\}}{x^2 - 4x + 4} = 5\), then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: The denominator (x-2)² creates a 0/0 form at x=2, so the numerator must also have (x-2) as a factor; factor the quadratic and use the limit definition to find k.
<p><strong>Step 1:</strong> Rewrite the denominator: x² - 4x + 4 = (x-2)²</p><p><strong>Step 2:</strong> For the limit to exist (0/0 form), the numerator must have (x-2) as a factor. Factor the quadratic: x² + (k-2)x - 2k = (x-2)(x+k)</p><p><strong>Step 3:</strong> Substitute this factorization:</p><p>$$\lim_{x \to 2} \frac{\tan(x-2)(x-2)(x+k)}{(x-2)^2} = \lim_{x \to 2} \frac{\tan(x-2)(x+k)}{x-2}$$</p><p><strong>Step 4:</strong> Use the standard limit $\lim_{u \to 0} \frac{\tan u}{u} = 1$, where u = (x-2):</p><p>$$\lim_{x \to 2} \frac{\tan(x-2)}{x-2} \cdot (x+k) = 1 \cdot (2+k) = 5$$</p><p><strong>Step 5:</strong> Solve for k: 2 + k = 5 → k = 3</p><p>∴ Answer: D (k = 3)</p>
Correct Answer: D