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

$\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + a}}{{x + b}}} \right)^{x + b}} = $

$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + \tan x}}{{1 + \sin x}}} \right)^{\text{cosec } x}}$ is equal to

If $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ and $\lim _{x \rightarrow 0^{-}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=l$,then which of the following is true?

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to