If $f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q} \\ -x, & \text{if } x \in \mathbb{Q}^c \end{cases}$,then $\lim_{x \to 0} f(x)$ is

The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right)$ is equal to

$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to

Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x$. Then, $\lim _{n \rightarrow \infty} \frac{[n \sqrt{2}]}{n}$ is equal to

Let $f : [1, 3] \to R$ be a function satisfying $\frac{x}{[x]} \le f(x) \le \sqrt{6 - x}$ for all $x \ne 2$ and $f(2) = 1$,where $R$ is the set of all real numbers and $[x]$ denotes the greatest integer function.
Statement $1$: $\lim_{x \to 2^-} f(x)$ exists.
Statement $2$: $f$ is continuous at $x = 2$.