The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

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

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:

$\operatorname{Lim}_{n}$ ${\rightarrow \infty} \left\{ \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right) \left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \dots \left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right) \right\}$ is equal to

Define a sequence $\{s_n\}$ of real numbers by $s_n = \sum_{k=0}^n \frac{1}{\sqrt{n^2+k}}$,for $n \geq 1$. Then,$\lim_{n \rightarrow \infty} s_n$:

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