Limits, Continuity & Differentiability
Discontinuity at Boundary
Grade 12
Question:
<p>Let $f(x) = \frac{\sqrt{2 + \cos x}}{8x - 4\sqrt{x}}$, $g(x) = \frac{2\cos x - \sin 2x}{e^{2x} - 1}$ (where $0 + 2x$ in denominator), and $h(x) = \begin{cases} f(x) & \text{for } x < \pi/2 \\ g(x) & \text{for } x > \pi/2 \end{cases}$. Which of the following does not hold?</p>
<p>(a) $h$ is continuous at $x = \pi/2$</p>
<p>(b) $h$ has an irremovable discontinuity at $x = \pi/2$</p>
<p>(c) [incomplete in source]</p>
Step-by-Step Solution
Key Concept: A function cannot be both continuous and have an irremovable discontinuity at the same point. Check one-sided limits.
<p><strong>Analysis:</strong> The question text appears incomplete in the source, but based on the structure:</p><p>Statements (a) and (b) are contradictory. Either $h$ is continuous at $\pi/2$ OR it has irremovable discontinuity there, not both.</p><p>To determine which statement does not hold, one must compute $\lim_{x \to (\pi/2)^-} f(x)$ and $\lim_{x \to (\pi/2)^+} g(x)$ and compare.</p><p>If these limits are equal, then (a) holds and (b) does not hold.</p><p>∴ Answer is likely (b).</p>
Correct Answer: B