$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$ is equal to

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

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ is equal to

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:

Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5$. Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = $