$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2} = $

The value of $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 2}}} \right)^{\frac{{x + 1}}{3}}}$ is equal to

$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)=$

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

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:

$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}} = $