If the function $f(x)$ satisfies $\mathop {\lim }\limits_{x \to 1} \frac{f(x)-2}{x^{2}-1}=\pi,$ evaluate $\mathop {\lim }\limits_{x \to 1} f(x).$

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

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ 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

If $\mathop {\lim }\limits_{x \to \infty } \left\{ {\ln \left( {{x^2} + 5x} \right) - 2\ln \left( {cx + 1} \right)} \right\} = -2$,then:

If $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$,then $P=$