If the function $f\left(x\right)=\left\{\begin{array}{r}{\left(\cos x\right)}^{\frac{1}{x}}; \text{if} x\ne 0\\ k ; \text{if} x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is 

Let $f\left(x\right)=\left|x\right|+|x-1|$, then 

The value of $f \left(0\right),$ so that the function $f\left(x\right)=\frac{(27-2x{)}^{{\displaystyle \frac{1}{3}}}-3}{9-3(243+5x{)}^{{\displaystyle \frac{1}{5}}}}\left(x\ne 0\right)$ is continuous, is given by 

The value of $f \left(0\right),$ so that the function $f\left(x\right)=\frac{2-(256-7x{)}^{{\displaystyle \frac{1}{8}}}}{(5x+32{)}^{{\displaystyle \frac{1}{5}}}-2},x\ne 0$ is continuous, is given by