Limits, Continuity & Differentiability
Indeterminate Forms [1^∞]
Grade 12
Question:
<p>The value of \(\lim_{x \to 7/2} \frac{\cot(2x - 7)}{2x^2 - 9x + 8}\) is equal to</p>
<p>(a) \(e^{5/2}\)</p>
<p>(b) \(e^{-5/2}\)</p>
<p>(c) \(e^{7/2}\)</p>
<p>(d) \(e^{3/2}\)</p>
Step-by-Step Solution
Key Concept: Convert the indeterminate form to exponential and evaluate the exponent limit using algebraic manipulation and trigonometric limits.
<p><strong>Step 1:</strong> Recognize the form $[1^\infty]$ as $x \to 7/2$.</p><p><strong>Step 2:</strong> Rewrite as $\lim_{x \to 7/2} \left\{1 + (2x^2 - 9x + 8)\right\}^{\cot(2x-7)}$</p><p><strong>Step 3:</strong> Evaluate $\lim_{x \to 7/2} (2x^2 - 9x + 8) \cdot \cot(2x-7)$</p><p><strong>Step 4:</strong> Factor: $2x^2 - 9x + 8 = (2x-1)(x-8)$ and use $\lim_{u \to 0} \frac{\cot u}{u} = \frac{1}{u}$ behavior.</p><p><strong>Step 5:</strong> The exponent limit equals $\frac{4x-9}{2}$ evaluated at $x = 7/2$, giving $\frac{5}{2}$.</p><p>∴ Answer is (a) $e^{5/2}$</p>
Correct Answer: A