If $\mathop {\lim }\limits_{x \to a} \frac{{{x^9} + {a^9}}}{{x + a}} = 9$,then $a = $

Consider the following statements:
Statement $1$: $\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = 1$ (where $a+b+c \neq 0$).
Statement $2$: $\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \frac{1}{4}$.

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2} = $

$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=$

$\lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n=$