$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^5} - 1}}{{{{(1 + x)}^3} - 1}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

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 $P =$

Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If $L$ and $R$ respectively denote the left-hand limit and the right-hand limit of $f(x)$ at $x=0$,then $\frac{32}{\pi^2}\left(L^2+R^2\right)$ is equal to:

Evaluate the limit: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$