Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(\lim_{x \to 1} \dfrac{x^4 - 1}{x - 1} = \lim_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}\), then \(k\) is __________ (up to two decimal places).</p>
Step-by-Step Solution
Key Concept: Both limits represent derivatives: the first is f'(1) where f(x)=x⁴, and the second is the ratio of derivatives at x=k. Use L'Hôpital's rule or algebraic simplification to find each limit, then equate them.
<p><strong>Step 1:</strong> Evaluate the first limit by factoring.</p><p>$$\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to 1} \frac{(x^2-1)(x^2+1)}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{x-1}$$</p><p>$$= \lim_{x \to 1} (x+1)(x^2+1) = (2)(2) = 4$$</p><p><strong>Step 2:</strong> Evaluate the second limit by factoring.</p><p>$$\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}$$</p><p>$$= \lim_{x \to k} \frac{x^2+xk+k^2}{x+k} = \frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}$$</p><p><strong>Step 3:</strong> Equate the two limits.</p><p>$$\frac{3k}{2} = 4$$</p><p>$$3k = 8$$</p><p>$$k = \frac{8}{3} = 2.67$$</p><p>∴ Answer: <strong>2.67</strong></p>
Correct Answer: 2