<p>We have \(\lim_{x \to 1} \dfrac{x^4 - 1}{x - 1} = \lim_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}\). Find the value of \(k\).</p>
Step-by-Step Solution
Key Concept: Factor both limits using algebraic identities (difference of powers) and use the fact that the limits are equal to find k. The left side equals 4, so the right side must also equal 4.
<p><strong>Step 1:</strong> Evaluate the left side using factorization.</p><p>$$\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{x-1} = \lim_{x \to 1} (x+1)(x^2+1)$$</p><p>$$= (1+1)(1^2+1) = (2)(2) = 4$$</p><p><strong>Step 2:</strong> Evaluate the right side and set it equal to 4.</p><p>$$\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)} = \lim_{x \to k} \frac{x^2+xk+k^2}{x+k}$$</p><p>$$= \frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}$$</p><p><strong>Step 3:</strong> Equate both limits.</p><p>$$\frac{3k}{2} = 4 \implies 3k = 8 \implies k = \frac{8}{3} = 2.67$$</p><p>∴ Answer: <strong>2.67</strong> or <strong>8/3</strong></p>
Correct Answer: 2.67