Limits, Continuity & Differentiability
Continuity of functions
Grade 12
Question:
<p>Since <em>f</em>(<em>x</em>) is continuous in <span>\(\left[0, \dfrac{\pi}{2}\right]\)</span>, find the value of <span>\(f\left(\dfrac{\pi}{4}\right) = \lim_{x \to \pi/4} f(x) = \lim_{x \to \pi/4} \left(\dfrac{1 - \tan x}{4x - \pi}\right)\)</span>. What is this limit equal to?</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(-\dfrac{1}{2}\)</p>
<p>\(1\)</p>
<p>\(-1\)</p>
Step-by-Step Solution
Key Concept: Since f(x) is continuous at π/4, direct substitution would give 0/0 (indeterminate form), requiring L'Hôpital's rule or algebraic manipulation to find the limit.
<p><strong>Step 1:</strong> Check the form at x = π/4. When x → π/4: numerator = 1 - tan(π/4) = 1 - 1 = 0 and denominator = 4(π/4) - π = π - π = 0. This is 0/0 form (indeterminate).</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule. Differentiate numerator and denominator separately:</p><p>• Numerator: d/dx(1 - tan x) = -sec²x</p><p>• Denominator: d/dx(4x - π) = 4</p><p><strong>Step 3:</strong> Evaluate the limit:</p><p>$$\lim_{x \to \pi/4} \frac{1 - \tan x}{4x - \pi} = \lim_{x \to \pi/4} \frac{-\sec^2 x}{4}$$</p><p><strong>Step 4:</strong> Substitute x = π/4:</p><p>$$= \frac{-\sec^2(\pi/4)}{4} = \frac{-2}{4} = -\frac{1}{2}$$</p><p>Since sec(π/4) = √2, we have sec²(π/4) = 2</p><p>∴ Answer: B (which equals -1/2)</p>
Correct Answer: B