$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$

If $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$ and $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$ are the roots of the equation $ax^{2} + bx - 4 = 0$,then the ordered pair $(a, b)$ is:

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} - 1}}{x} = $

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}$ is equal to