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

If $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$,then the value of $\frac{\log _e \alpha}{1+\log _e \alpha}$ equals :

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = $

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{A+e^{n x}}{x+A e^{n x}}$

$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)

$\lim _{z \rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1} = $