Let $f(x) = \begin{cases} x^2 + k, & \text{when } x \ge 0 \\ -x^2 - k, & \text{when } x < 0 \end{cases}$. If the function $f(x)$ is continuous at $x = 0$,then $k =$

If the function $f(x) = \begin{cases} (\cos x)^{1/x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

If $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{\tan x \cdot \log(1+x)}$ for $x \neq 0$, is continuous at $x = 0$, then the value of $f(0)$ is equal to ...

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{for } -1 \leq x < 0 \\ 2x^2+3x-2, & \text{for } 0 \leq x \leq 1 \end{cases}$ is continuous at $x=0$,then $k$ is equal to

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.