If $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then

The function $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$ is

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 the function given by $f(x) = \left(\frac{4x+1}{1-4x}\right)^{\frac{1}{x}}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.