Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{\theta \to \pi/4} \frac{\sqrt{2} - \cos\theta - \sin\theta}{(4\theta - \pi)^2}\) is equal to</p>
<p>(a) \(\frac{\sqrt{2}}{4}\)</p>
<p>(b) \(\frac{\sqrt{2}}{8}\)</p>
<p>(c) \(\frac{\sqrt{2}}{16}\)</p>
<p>(d) \(\frac{\sqrt{2}}{32}\)</p>
Step-by-Step Solution
Key Concept: At θ = π/4, both numerator and denominator equal zero, requiring L'Hôpital's rule or Taylor expansion around θ = π/4. Use substitution h = θ - π/4 to simplify the expansion of trigonometric functions.
<p><strong>Step 1:</strong> Check form at θ = π/4: numerator = √2 - cos(π/4) - sin(π/4) = √2 - √2/2 - √2/2 = 0, denominator = 0. This is 0/0 form.</p><p><strong>Step 2:</strong> Let h = θ - π/4, so θ = π/4 + h and 4θ - π = π + 4h - π = 4h. The limit becomes:<br/>$$\lim_{h \to 0} \frac{\sqrt{2} - \cos(\frac{\pi}{4}+h) - \sin(\frac{\pi}{4}+h)}{16h^2}$$</p><p><strong>Step 3:</strong> Expand using angle addition formulas:<br/>$$\cos(\frac{\pi}{4}+h) = \frac{\sqrt{2}}{2}\cos h - \frac{\sqrt{2}}{2}\sin h = \frac{\sqrt{2}}{2}(1 - \frac{h^2}{2} + ...) - \frac{\sqrt{2}}{2}(h - ...)$$<br/>$$\sin(\frac{\pi}{4}+h) = \frac{\sqrt{2}}{2}\sin h + \frac{\sqrt{2}}{2}\cos h = \frac{\sqrt{2}}{2}(h + ...) + \frac{\sqrt{2}}{2}(1 - \frac{h^2}{2} + ...)$$</p><p><strong>Step 4:</strong> Numerator = $$\sqrt{2} - \frac{\sqrt{2}}{2}(2 - \frac{h^2}{2} - 2h + ...) = \sqrt{2} - \sqrt{2} + \frac{\sqrt{2}h^2}{4} + ... = \frac{\sqrt{2}h^2}{4} + O(h^3)$$</p><p><strong>Step 5:</strong> Therefore:<br/>$$\lim_{h \to 0} \frac{\frac{\sqrt{2}h^2}{4}}{16h^2} = \frac{\sqrt{2}}{64}$$</p><p>∴ Answer: D</p>
Correct Answer: D