Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>The value of \(k\) for which the function \[f(x) = \begin{cases} \left(\dfrac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \dfrac{\pi}{2} \\ k + \dfrac{2}{5}, & x = \dfrac{\pi}{2} \end{cases}\] is continuous at \(x = \dfrac{\pi}{2}\) is</p>
<p>\(\dfrac{17}{20}\)</p>
<p>\(\dfrac{3}{5}\)</p>
<p>\(-\dfrac{2}{5}\)</p>
<p>\(\dfrac{2}{5}\)</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, evaluate the limit of the exponent as x→0 using L'Hôpital's rule: tan(4x)/tan(5x) → 4/5, so f(0⁻) = (4/5)^(4/5). For f to be continuous, k must equal this limit value.
<p><strong>Step 1:</strong> Find the left-hand limit as x → 0⁻</p><p>We need: lim(x→0⁻) (4/5)^(tan(4x)/tan(5x))</p><p><strong>Step 2:</strong> Evaluate the exponent limit using L'Hôpital's rule</p><p>lim(x→0) tan(4x)/tan(5x) = lim(x→0) [4sec²(4x)]/[5sec²(5x)] = 4/5</p><p><strong>Step 3:</strong> Substitute back into the limit</p><p>lim(x→0⁻) f(x) = (4/5)^(4/5)</p><p><strong>Step 4:</strong> Apply continuity condition</p><p>For f to be continuous at x = 0: f(0) = lim(x→0⁻) f(x)</p><p>Therefore: k = (4/5)^(4/5)</p><p>∴ Answer: D</p>
Correct Answer: D