Limits, Continuity & Differentiability
Evaluation of Limits (L'Hospital's Rule)
Grade None
Question:
<p>Evaluate \(\lim_{x \to \pi/2} \tan x \cdot \log \sin x\).</p>
Step-by-Step Solution
Key Concept: Rewrite the indeterminate form ∞·(-∞) as a fraction to apply L'Hôpital's rule. Convert tan x · log sin x = (log sin x)/(cot x) to get 0/0 form as x → π/2.
<p><strong>Step 1:</strong> Identify the form. As x → π/2⁻: tan x → ∞ and log sin x → log(1) = 0, giving 0·∞ indeterminate form.</p><p><strong>Step 2:</strong> Rewrite as a fraction: tan x · log sin x = (log sin x)/(cot x).</p><p><strong>Step 3:</strong> As x → π/2⁻: numerator → 0 and denominator → 0, so we have 0/0 form. Apply L'Hôpital's rule.</p><p><strong>Step 4:</strong> Differentiate: d/dx(log sin x) = (cos x)/(sin x) = cot x and d/dx(cot x) = -csc² x.</p><p><strong>Step 5:</strong> Apply L'Hôpital: lim(x→π/2⁻) [cot x]/[-csc² x] = lim(x→π/2⁻) [cot x · sin² x]/[-1] = lim(x→π/2⁻) [cos x · sin x]/[-1].</p><p><strong>Step 6:</strong> Substitute x = π/2: [cos(π/2) · sin(π/2)]/[-1] = [0 · 1]/[-1] = 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0