$\mathop {\lim }\limits_{x \to 0} \frac{|x|}{x} = $

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

If $\alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ and $\beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$,then

$\mathop {\lim }\limits_{x \to 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} \right] = $

If $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}} = 2b \sin^2 \theta$,where $b > 0$ and $\theta \in (-\pi, \pi]$,then the value of $\theta$ is

$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$