If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)

Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & \text{if } x>0 \\ 2, & \text{if } x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text{if } x < 0 \end{cases}$ where $[x]$ denotes the greatest integer function. If $f$ is continuous at $x=0$,then $\beta-\alpha$ is equal to

Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ is:

If $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then:

The function $f(x) = \begin{cases} x - 1, & x < 2 \\ 2x - 3, & x \ge 2 \end{cases}$ is a continuous function: