The value of $\mathop {\lim }\limits_{x \to 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$ is

$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta (\tan \theta - \sin \theta )}}{{{{(1 - \cos 2\theta )}^2}}}$ is

$\lim _{x}$ ${\rightarrow 0} \left( \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \right) \left( \frac{\sin ^3(4 x)}{(\log _e(2 x+1))^5} \right) \right)$ is equal to $.........$.

If $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$,then $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = $

$\lim _{x \rightarrow 0} \frac{x \tan 4x - 2x \tan 2x}{(1 - \cos 4x)^2} = $

$\lim _{x \rightarrow 0} \frac{\tan ^4 x-\sin ^4 x}{x^6} = $