Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x}\) is equal to</p>
<p>(a) 1</p>
<p>(b) \(-2\)</p>
<p>(c) \(\frac{1}{2}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Recognize this as a 0/0 indeterminate form that requires L'Hôpital's rule or algebraic manipulation. The numerator approaches 0 as x → π/2, and the denominator also approaches 0.
<p><strong>Step 1:</strong> Check the form. As x → π/2: numerator → 2(π/2) - π = 0 and denominator → cos(π/2) = 0. This is 0/0 form.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule: differentiate numerator and denominator separately.</p><p>$$\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} = \lim_{x \to \pi/2} \frac{\frac{d}{dx}(2x-\pi)}{\frac{d}{dx}(\cos x)} = \lim_{x \to \pi/2} \frac{2}{-\sin x}$$</p><p><strong>Step 3:</strong> Substitute x = π/2: $$\frac{2}{-\sin(\pi/2)} = \frac{2}{-1} = -2$$</p><p>∴ Answer: <strong>-2</strong></p>
Correct Answer: B