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

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

$\lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x} - \lim _{x \rightarrow 0} \frac{\log (1+x^3)}{\sin ^3 x} =$

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}$ for $n \geq 2$. Then,the value of $\lim _{n \rightarrow \infty} x_{n}$ is

Find $\mathop {\lim }\limits_{x \to 5} f(x),$ where $f(x)=|x|-5$.