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

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}} \right)$ equals

The value of $\lim_{x \to 0} (\frac{8}{x^8}) [1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cdot \cos \frac{x^2}{4}]$ is equal to:

If $a, b, c$ and $d$ are positive,then $\lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$ is equal to:

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is