Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Let <span>\( f(x) = \dfrac{1 - \tan x}{4x - \pi} \)</span>, <span>\( x \neq \dfrac{\pi}{4} \)</span>, <span>\( x \in \left[0, \dfrac{\pi}{2}\right] \)</span>. If <span>\( f(x) \)</span> is continuous in <span>\( \left[0, \dfrac{\pi}{2}\right] \)</span>, then <span>\( f\!\left(\dfrac{\pi}{4}\right) \)</span> is</p>
<p>1</p>
<p>\( \dfrac{1}{2} \)</p>
<p>\( -\dfrac{1}{2} \)</p>
<p>\( -1 \)</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous at x = π/4, we must find lim[x→π/4] f(x) using L'Hôpital's rule since both numerator and denominator approach 0 at x = π/4. The value f(π/4) must equal this limit.
<p><strong>Step 1:</strong> Check the form at x = π/4.</p><p>At x = π/4: numerator = 1 - tan(π/4) = 1 - 1 = 0, denominator = 4(π/4) - π = 0</p><p>This is a 0/0 indeterminate form, so L'Hôpital's rule applies.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule.</p><p>lim[x→π/4] (1 - tan x)/(4x - π) = lim[x→π/4] d/dx(1 - tan x) / d/dx(4x - π)</p><p>= lim[x→π/4] (-sec²x) / 4</p><p><strong>Step 3:</strong> Evaluate at x = π/4.</p><p>= (-sec²(π/4)) / 4</p><p>= (-[√2]²) / 4</p><p>= -2/4</p><p>= -1/2</p><p><strong>Step 4:</strong> By continuity at x = π/4.</p><p>f(π/4) = lim[x→π/4] f(x) = -1/2</p><p>∴ Answer: C (which is -1/2)</p>
Correct Answer: C