Q.Select the correct answer from the given alternatives. $f(x) = \dfrac{(16^x-1)(9^x-1)}{(27^x-1)(32^x-1)}$, for $x \ne 0$, $= k$, for $x=0$, is continuous at $x=0$, then $k=$ (A) $\dfrac83$ (B) $\dfrac{8}{15}$ (C) $-\dfrac{8}{15}$ (D) $\dfrac{20}{3}$

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 rig …