Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{(x+1)^{5}-1}{x}$

If $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$, then $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ is

If $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$,then $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = $

$\mathop {\lim }\limits_{x \to \pi /2} \frac{\tan 3x}{x} = $

The value of the limit of $\frac{x^3 - 8}{x^2 - 4}$ as $x \to 2$ is

The value of $\lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$ is