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

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 $\lim _{x \rightarrow \infty} \frac{f(6 x)}{f(3 x)}=$

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

If $\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}}=k$ and $\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=l$,then $k+l=$

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 $f : (1, 2) \to R$ 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