Question:

<p>A removable discontinuity exists at a point if $\lim_{x \to a} f(x)$ exists but $f(a)$ is either undefined or $f(a) \neq \lim_{x \to a} f(x)$.</p><p><strong>(a)</strong> For $f(x) = \frac{1}{1 + 2\cot x}$: As $x \to 0$, $\cot x \to \infty$, so $\lim_{x \to 0} f(x) = 0$. This is a removable discontinuity. ✓</p><p><strong>(b)</strong> For $f(x) = \cos\left(\frac{\pi}{|\sin x|}\right)$: As $x \to 0$, $|\sin x| \to 0^+$, so $\frac{\pi}{|\sin x|} \to +\infty$. The limit does not exist. Not removable. ✗</p><p><strong>(c)</strong> For $f(x) = x\sin\left(\frac{1}{x}\right)$: Since $|\sin(1/x)| \leq 1$, we have $\lim_{x \to 0} x\sin(1/x) = 0$ (squeeze theorem). This is removable. ✓</p><p><strong>(d)</strong> For $f(x) = \frac{|\sin x|}{\ln|x|}$: As $x \to 0$, numerator $\to 0$ but denominator $\to -\infty$. Thus limit $= 0$, but $\ln|x|$ undefined at $x=0$ makes analysis complex. Not typically removable. ✗</p>