Is $\lim_{x\to\infty} (0.99999...)^x=1$?

Interpreting $0.9999\ldots$ as $\lim_{n\to\infty}(1-{1\over10^n})$, which is clearly equal to $1$, what we have is a case where limits do not interchange:

$$1=\lim_{x\to\infty}1^x=\lim_{x\to\infty}\left(\lim_{n\to\infty}\left(1-{1\over10^n}\right)\right)^x\not=\lim_{n\to\infty}\left(\lim_{x\to\infty}\left(1-{1\over10^n}\right)^x\right)=\lim_{n\to\infty}0=0$$

Yes, if you mean $0.\overline{9}$ by $0.999\ldots$ then it is true, since $$\lim_{x\to \infty} 1^x = 1$$ and, as you said, $0.\overline{9}$ , is just another way to write 1.

If you just mean a lot of nine's, then it is not true, and the limit would go to $0$

If $x=y$, any truth about $x$ is a truth about $y$—both $x$ and $y$ represent the same object. Sometimes the notation we use can distract us from this fact. As Kinheadpump has already noted, since $$ \lim_{x \to \infty}1^x = 1 \, , $$ it is also true that $$ \lim_{x \to \infty}(0.\overline{9})^x=1 \, . $$ Indeed, this is actually the same statement repeated over.

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Keep in mind that $0.999...$ has a precise meaning. It refers to a limit of a series: $$ \lim_{N \to \infty}\sum_{n=1}^{N}\frac{9}{10^n} = 0.9 + 0.09 + 0.009 + \ldots = 1 \, . $$ If we instead consider the series $$ S=\sum_{n=1}^{N}\frac{9}{10^n} $$ for some finite value of $N$, then $$ \lim_{x \to \infty}S^x = 0 $$ Hence, the reason that $\lim_{x \to \infty}(0.\overline{9})^x=1$ is because $0.999...$ refers to the limit of a series, rather than the sum of finitely many terms. And since the limit of this series is $1$, it follows that $0.\overline{9}=1$. You are correct in saying that $0.999...$ is merely an alternative symbol for $1$, just as $5+7$ is an alternative way of writing $12$.