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

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

The value of $\lim _{x \rightarrow 0^{+}} \frac{x}{p} \left[ \frac{q}{x} \right]$ is

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

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

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to: