If $\mathop {\lim }\limits_{x \to 2} \frac{{\tan (x - 2)({x^2} + (a - 2)x - 2a)}}{{({x^2} - 4x + 4)}} = 7$,then $a$ is -

If $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}$ exists and is equal to $2$,then the ordered pair $(\alpha, \beta)$ of real numbers is

Let $\alpha, \beta \in R$ be such that $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$. Then $6(\alpha+\beta)$ equals

If $\lim_{x \to \infty} [\frac{x^2 + x + 1}{x + 1} - ax - b] = 3$, then $a - b = $

If $\lim _{x \rightarrow 0} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0, (n \neq 0)$,then the minimum possible positive value of $a$ is

The product of all possible values of $\alpha$, for which $\lim_{x \to 0} \left( \frac{1 - \cos(\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2$, is: