In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = [x \sin \pi x]$,then $f(x)$ is

Let a function $f: R \rightarrow R$ be defined as
$f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}$
where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$,then $(a + b)$ is equal to:

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is: