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

The value of the limit $\lim_{x \rightarrow 1} \frac{\sin(e^{x-1}-1)}{\log x}$ is

$\lim _{n}$ ${\rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots +\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots +n^3\right)-\left(1^2+2^2+\ldots +n^2\right)}$ is equal to:

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $

The value of the limit $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$ is