Q.Select the correct answer from the given alternatives. $f(x) = \dfrac{2^{\cot x}-1}{\pi-2x}$, for $x \ne \dfrac{\pi}{2}$, $= \log\sqrt2$, for $x = \dfrac{\pi}{2}$. (A) $f$ is continuous at $x=\dfrac{\pi}{2}$ (B) $f$ has a jump discontinuity at $x=\dfrac{\pi}{2}$ (C) $f$ has a removable discontinuity (D) $\lim_{x\to\frac{\pi}{2}} f(x) = 2\log 3$

A function $f(x)$ is continuous at a point $x=a$ when three separate requirements all hold together: the function must actually be defined at $a$ (and on a whole open interval around it), the limit $\lim_{x\to a} f(x)$ must exist as a single finite number, and that limiting value must be exactly equal to $f(a)$ itself. If any one of these three fails — the value is missing, the two-sided limit does not exist, or the limit disagrees with the assigned value — the function is discontinuous at that point. An equivalent way to state the third condition is $\lim_{h\to0}[f(a+h)-f(a)]=0$, which says that as the input is nudged away from $a$ by a shrinking amount $h$, the resulting change in output must also shrink to zero, ruling out any sudden jump right at $a$. In practice, checking continuity at a point almost always comes down to computing $f(a)$ directly from its formula, computing the two-sided limit (often by simplifying a $0/0$ form through factoring, rationalising, or a standard limit such as $\lim_{x\to0}\frac{\sin x}{x}=1$), and then comparing the two numbers.