$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{2x\tan x - x\tan 2x}}$ is

If $f(x) = \frac{x(a^x - 1)}{1 - \cos x}$ and $g(x) = \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$,then $\lim_{x \to 0} (f(x) - g(x)) = $

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

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is

If $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$ and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$,then $\lim_{x \to 3} [f(x) + g(x) + h(x)]$ is