$A$ point in the domain of a function where the discontinuity cannot be removed by redefining the function at that point is called:

Let $f(x) = \min \{x, x^2\}$ for every real number $x$. Then:

Consider the function $f(x) = \begin{cases} \frac{x}{[x]} & \text{if } 1 \leqslant x < 2 \\ 1 & \text{if } x = 2 \\ \sqrt{6-x} & \text{if } 2 < x \leqslant 3 \end{cases}$ where $[x]$ denotes the greatest integer function. At $x = 2$,the function:

Let $[\bullet]$ denote the greatest integer function,and let $f(x) = \min \{\sqrt{2}x, x^2\}$. Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$. Then $\sum_{x \in S} f(x)$ equals:

If $\begin{aligned} f(x) &= \frac{4 \sin \pi x}{5 x} \text{ for } x \neq 0 \\ &= 2k \text{ for } x = 0 \end{aligned}$ is continuous at $x = 0$,then the value of $k$ is

If the function $f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at $x = 0$,then $f(0)$ is equal to . . . . . . .