Let $f: R \rightarrow (0, \infty)$ be a strictly increasing function such that $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$. Then,the value of $\lim _{x \rightarrow \infty} \left[\frac{f(5 x)}{f(x)}-1\right]$ is equal to

Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x$. Then, $\lim _{n \rightarrow \infty} \frac{[n \sqrt{2}]}{n}$ is equal to

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

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

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