If $f(x) = \begin{cases} \frac{1}{x} \sin(x^2), & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

For every integer $n$,let $a_n$ and $b_n$ be real numbers. Let function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \begin{cases} a_n + \sin \pi x, & \text{for } x \in [2n, 2n+1] \\ b_n + \cos \pi x, & \text{for } x \in (2n-1, 2n) \end{cases}$,for all integers $n$. If $f$ is continuous,then which of the following hold$(s)$ for all $n$?

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

Let $f$ and $g$ be real-valued functions. If $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{[f(x)+7]^{2 / 3}}=\frac{7}{4}$, $\lim _{x \rightarrow 0} f(x)=1$ and $\lim _{x \rightarrow 0} g(x)=\alpha$, then $h(x)= \begin{cases} \sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5} \end{cases}$ is:

$A$ point in the domain of a function where the discontinuity cannot be removed by redefining the function at that point is called:

If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is