<p>Find \(\displaystyle\lim_{x \to \pi/2} \frac{\left[1 - \tan\left(\dfrac{x}{2}\right)\right]\left[1 - \sin x\right]}{\left[1 + \tan\left(\dfrac{x}{2}\right)\right]\left[\pi - 2x\right]^3}\).</p>
Step-by-Step Solution
Key Concept: Use the substitution u = π/2 - x to convert the limit to u → 0, then apply Taylor expansions for tan(u/2) and sin(π/2 - u) around u = 0 to resolve the 0/0 indeterminate form.
<p><strong>Step 1: Substitution</strong> Let u = π/2 - x, so x = π/2 - u and as x → π/2, u → 0. Then π - 2x = 2u and x/2 = π/4 - u/2.</p><p><strong>Step 2: Rewrite the expression</strong></p><p>$$\frac{[1 - \tan(π/4 - u/2)][1 - \sin(π/2 - u)]}{[1 + \tan(π/4 - u/2)](2u)^3}$$</p><p><strong>Step 3: Apply Taylor expansions around u = 0</strong></p><p>$$\tan(π/4 - u/2) = \tan(π/4)\frac{1 + \tan(π/4)·(u/2)}{1 - \tan(π/4)·(u/2)} = \frac{1 + u/2}{1 - u/2} ≈ 1 + u + O(u²)$$</p><p>$$\sin(π/2 - u) = \cos(u) ≈ 1 - \frac{u^2}{2} + O(u^4)$$</p><p><strong>Step 4: Substitute expansions</strong></p><p>Numerator: $[1 - (1 + u + O(u²))][1 - (1 - u²/2)] = [-u + O(u²)][u²/2 + O(u³)] ≈ -\frac{u^3}{2}$</p><p>Denominator: $[1 + (1 + u)][2u]^3 = (2 + u)·8u^3 ≈ 16u^3$ (leading term)</p><p><strong>Step 5: Evaluate the limit</strong></p><p>$$\lim_{u \to 0} \frac{-u^3/2}{16u^3} = -\frac{1}{32}$$</p><p>∴ Answer: C</p>
Correct Answer: C