If $[\cdot]$ denotes the greatest integer function,then $\lim _{x \rightarrow \frac{-3}{5}} \frac{1}{x}\left[\frac{-1}{x}\right]=$

For $A \neq 0$ and $x < 0$,evaluate $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

$\lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ is equal to :