If the function $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \ne 1 \\ k, & x = 1 \end{cases}$ is continuous at $x = 1$,then the value of $k$ is:

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

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le 5 \\ 3x - 5, & \text{if } x > 5 \end{cases}$ at $x = 5$. (in $/5$)

The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ for all $x \in R$. Consider the following statements:
$I.$ $f$ is an odd function.
$II.$ $f$ is an even function.
$III.$ $f$ is differentiable everywhere.
Then,

Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$