Limits, Continuity & Differentiability
Limits involving floor function
Grade 12
Question:
<p><strong>Ex. 40 (D):</strong> Let <span class="math">\tan\left(\frac{2}{3}|\sin\theta|\right) = \cot\left(\frac{2}{3}|\cos\theta|\right)\</span>, where <span class="math">\theta \in \mathbb{R}\</span> and <span class="math">f(x) = \left(|\sin\theta| + |\cos\theta|\right)^x\.</span> The value of <span class="math">\lim_{x \to \infty} \left[\frac{4 \cdot 2^7}{56 \cdot f(x)}\right]\</span> equals (where [·] represents the greatest integer function)</p>
Step-by-Step Solution
Key Concept: Use the complementary angle relationship and bounds on trigonometric sums to find the base of the exponential.
<p><strong>Solution:</strong> From the given equation:</p><p>\[\tan\left(\frac{2}{3}|\sin\theta|\right) = \cot\left(\frac{2}{3}|\cos\theta|\right)\]</p><p>This implies: <span class="math">\frac{2}{3}|\sin\theta| + \frac{2}{3}|\cos\theta| = \frac{\pi}{2}\</span></p><p>Therefore: <span class="math">|\sin\theta| + |\cos\theta| = \frac{3\pi}{4}\</span></p><p>Since <span class="math">1 \leq |\sin\theta| + |\cos\theta| \leq \sqrt{2}\</span> and <span class="math">1 \leq \frac{3\pi}{4} \leq 2\</span>:</p><p>\[\lim_{x \to \infty} \left[\frac{4 \cdot 2^7}{56 \cdot f(x)}\right] = 0\]</p><p>∴ Answer is (r)</p>
Correct Answer: r