Limits, Continuity & Differentiability
Algebraic Method for Limits
Grade 12
Question:
<p>If \(\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}\), then \(k\) is</p>
<p>(a) \(\frac{3}{4}\)</p>
<p>(b) \(\frac{4}{3}\)</p>
<p>(c) \(\frac{3}{2}\)</p>
<p>(d) \(\frac{8}{3}\)</p>
Step-by-Step Solution
Key Concept: Use algebraic factorization to simplify rational expressions before taking limits. Evaluate both sides and equate them to find k.
<p><strong>Solution:</strong></p><p>Given: $\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}$</p><p>For the left side:</p><p>$\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{x-1} = \lim_{x \to 1} (x+1)(x^2+1) = 2 \cdot 2 = 4$</p><p>For the right side:</p><p>$\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{(x-k)(x^2 + k^2 + xk)}{(x-k)(x+k)} = \lim_{x \to k} \frac{x^2 + k^2 + xk}{x+k} = \frac{3k^2}{2k}$</p><p>Setting them equal:</p><p>$4 = \frac{3k^2}{2k}$</p><p>$8k = 3k^2$</p><p>$k = \frac{8}{3}$</p><p>∴ Answer is (d) $\frac{8}{3}$</p>
Correct Answer: D