Will it grow or decline?

The first thing is to prove that the notation $\displaystyle 0.\overline { 9 }$ is "exactly" 1.

Consider $\displaystyle \frac { 1 }{ 3 }= 0.\overline { 3 }$

$\displaystyle 3 \times \frac { 1 }{ 3 }= 3 \times 0.\overline { 3 } =3 \times 0.3333333333...=0.9999999999...=0.\overline { 9 }$

Since $\displaystyle 3 \times \frac { 1 }{ 3 }= 1$

$\displaystyle 3 \times \frac { 1 }{ 3 } =1= 0.\overline { 9 }$

Then

$\displaystyle \lim _{ x\rightarrow \infty }{ \left( 0.\overline { 9 } \right)^{ x } } = \lim _{ x\rightarrow \infty }{ \left( 1 \right)^{ x } }=1$

Let $\alpha=0.\overline{9}=\lim_{n\rightarrow\infty}\sum_{r=1}^n\frac{9}{10^r}=\lim_{n\rightarrow\infty}\left(1-10^{-n}\right)=1$

Then, required limit $\lim_{n\rightarrow\infty}x_n$ where $x_n=\alpha^n=1^n=1$ .

Hence, $x_n$ is a constant sequence. So, $\lim_{n\rightarrow\infty}x_n=\boxed{1}$