If $f$ is defined by $f(x) = \begin{cases} \frac{1-\cos ax}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x = 0 \end{cases}$ and $f$ is continuous at $x=0$,then $a^{2} =$ . . . . . . .

Let $f(x) = \begin{cases} (3 - \sin(1/x))|x|, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$f$ has a

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then

The function $f(x) = [x]^2 - [x^2]$,(where $[y]$ is the greatest integer less than or equal to $y$),is discontinuous at

Statement $1$: $A$ function $f: R \to R$ is continuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) = f(x_0)$.
Statement $2$: $A$ function $f: R \to R$ is discontinuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) \neq f(x_0)$.

Let $f$ be a differentiable function on the open interval $(a, b)$. Which of the following statements must be true?
$I$. $f$ is continuous on the closed interval $[a, b]$
$II$. $f$ is bounded on the open interval $(a, b)$
$III$. If $a < a_1 < b_1 < b$,and $f(a_1) < 0 < f(b_1)$,then there is a number $c$ such that $a_1 < c < b_1$ and $f(c) = 0$