Let $f: R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$. Then $\lim_{x \to \infty} \frac{f(2x)}{f(x)} = $

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 $x = 2$.

Let $f: R^{+} \rightarrow R$ be an increasing function such that $f(x) > 0$ for all $x$. If $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)}=1$,then $\lim _{x \rightarrow \infty} \frac{f(6 x)}{f(3 x)}=$

Let $f: (0, \infty) \rightarrow \mathbb{R}$ and $g: (0, \infty) \rightarrow \mathbb{R}$ be two functions where $g(x) = x + \frac{1}{x}$. If $1 < f(x) \cdot g(x) < 10$ for all $x > 0$,then $\lim_{x \to \infty} f(x)$ is

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

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