Let $f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$,and $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$,then $P =$

$\lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n=$

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-

The set of all values of $a$ for which $\lim_{x \rightarrow a}(\lfloor x-5 \rfloor - \lfloor 2x+2 \rfloor) = 0$,where $\lfloor \alpha \rfloor$ denotes the greatest integer less than or equal to $\alpha$,is equal to

$\lim _{x}$ ${\rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}$ is equal to

If $f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$,then $\lim_{x \to 0^-} f(x)$ is: