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 $f : [1, 3] \to R$ 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 $x = 2$.

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

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

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)} = $

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,$\lim_{n \rightarrow \infty} s_n$: