If $n > 0$ and $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0$,then the minimum value of $a$ 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 $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$,then $a b=$

For $t > -1$,let $\alpha_t$ and $\beta_t$ be the roots of the equation $\left((t+2)^{\frac{1}{7}}-1\right) x^2+\left((t+2)^{\frac{1}{6}}-1\right) x+\left((t+2)^{\frac{1}{21}}-1\right)=0$. If $\lim _{t \rightarrow -1^{+}} \alpha_t$ and $\lim _{t \rightarrow -1^{+}} \beta_t$ are the roots of the limiting equation,and $a+b$ is the sum of these roots,then $72(a+b)^2$ is equal to . . . . . . .

If $\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right] = 2$,then

Given $f(x) = \frac{ax + b}{x + 1}$,$\lim_{x \rightarrow \infty} f(x) = 1$ and $\lim_{x \rightarrow 0} f(x) = 2$,then $f(-2)$ is