$\lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} = $

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

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

If $f(x) = \begin{cases} x\sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$

Let $f : (1, 2) \to R$ satisfy the inequality $\frac{\cos(2x - 4) - 33}{2} < f(x) < \frac{x^2 |4x - 8|}{x - 2}$ for all $x \in (1, 2)$. Then $\lim_{x \to 2^-} f(x)$ 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 $x = 2$.