Find $\mathop {\lim }\limits_{x \to 1} f(x)$,where $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$

The quadratic equation whose roots are $m$ and $n$,where $m = \lim_{x \rightarrow 0} \frac{x \log(1+2x)}{x \tan x}$ and $n = \lim_{x \rightarrow 0} \frac{\log x + \log(\frac{1+x}{x})}{x}$,is

$\lim _{x \rightarrow \infty} \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}} = $

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

$\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}=$