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

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

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = $

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

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

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