Limits, Continuity & Differentiability
Limits of Trigonometric Functions
Grade 12
<p>\(\lim_{x \to \pi/2} \dfrac{\left[1 - \tan\left(\dfrac{x}{2}\right)\right][1 - \sin x]}{\left[1 + \tan\left(\dfrac{x}{2}\right)\right][\pi - 2x]^3}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that as x → π/2, we have π/2 - x → 0, and use the substitution h = π/2 - x to convert to a standard limit form. Express tan(x/2) and sin(x) in terms of h using half-angle identities to resolve the 0/0 indeterminate form.
<p><strong>Step 1: Substitute h = π/2 - x</strong></p><p>As x → π/2, h → 0. Then x = π/2 - h, so:</p><ul><li>x/2 = π/4 - h/2</li><li>sin x = sin(π/2 - h) = cos h</li><li>π - 2x = π - 2(π/2 - h) = 2h</li></ul><p><strong>Step 2: Rewrite the limit</strong></p><p>$$\lim_{h \to 0} \frac{[1 - \tan(π/4 - h/2)][1 - \cos h]}{[1 + \tan(π/4 - h/2)](2h)^3}$$</p><p><strong>Step 3: Use tan identity</strong></p><p>tan(π/4 - h/2) = (1 - tan(h/2))/(1 + tan(h/2))</p><p>Therefore:</p><ul><li>1 - tan(π/4 - h/2) = 2tan(h/2)/(1 + tan(h/2))</li><li>1 + tan(π/4 - h/2) = 2/(1 + tan(h/2))</li></ul><p><strong>Step 4: Substitute and simplify</strong></p><p>$$\lim_{h \to 0} \frac{[2\tan(h/2)/(1+\tan(h/2))][1-\cos h]}{[2/(1+\tan(h/2))](8h^3)}$$</p><p>$$= \lim_{h \to 0} \frac{\tan(h/2)(1-\cos h)}{8h^3}$$</p><p><strong>Step 5: Apply standard limits</strong></p><p>Using 1 - cos h = 2sin²(h/2), tan(h/2) ≈ h/2, sin(h/2) ≈ h/2 as h → 0:</p><p>$$= \lim_{h \to 0} \frac{(h/2) \cdot 2(h/2)^2}{8h^3} = \lim_{h \to 0} \frac{h^3/4}{8h^3} = \frac{1}{32}$$</p><p>∴ Answer: <strong>C (1/32)</strong></p>
Correct Answer: C