$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

Let $\alpha$ and $\beta$ be real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is $....$ (in $.40$)

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

$\lim _{x \rightarrow \pi / 6} \left[ \frac{3 \sin x - \sqrt{3} \cos x}{6x - \pi} \right]$ is equal to:

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$ is

If $f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2$,then the value of $\lim_{x \to a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$ is: