If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$, then find the value of $a^2+b^2$.

Is the function defined by $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$ a continuous function?

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 . . . . . . .

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...